CP Stoichiometry Practice Test Notes

CP Stoichiometry Practice Test

Multiple Choice Questions (NaHCO3 Decomposition)

• Question 1: The balanced equation for the decomposition of sodium bicarbonate (NaHCO3) is:
  $2 NaHCO3 → Na2CO3 + CO2 + H2O$
  The corresponding coefficients are 2:1:1:1. So the answer is B.
• Question 2: If 6 moles of NaHCO3 decompose, the theoretical number of moles of CO2 produced is:
  Based on the balanced equation, 2 moles of NaHCO3 produce 1 mole of CO2.
  $\frac{6 \text{ moles NaHCO3}}{2 \text{ moles NaHCO3/mole CO2}} = 3 \text{ moles CO2}$
  The answer is D.
• Question 3: If 8 moles of NaHCO3 decompose, the theoretical number of moles of H2O produced is:
  Based on the balanced equation, 2 moles of NaHCO3 produce 1 mole of H2O.
  $\frac{8 \text{ moles NaHCO3}}{2 \text{ moles NaHCO3/mole H2O}} = 4 \text{ moles H2O}$
  The answer is A.
• Question 4: The theoretical mass of CO2 produced from 4 moles of NaHCO3 is:
  From the balanced equation, 2 moles of NaHCO3 produce 1 mole of CO2. Therefore, 4 moles of NaHCO3 produce 2 moles of CO2.
  The molar mass of CO2 is approximately 44 g/mol.
  $2 \text{ moles CO2} \times 44 \frac{\text{g}}{\text{mole}} = 88 \text{ g}$
  The answer is D.
• Question 5: The % yield of an experiment using 1 mole of NaHCO3 and isolating 22 g of CO2 is:
  From the balanced equation, 2 moles NaHCO3 should produce 1 mole CO2. Thus, 1 mole NaHCO3 should theoretically produce 0.5 moles CO2.
  The molar mass of CO2 is 44 g/mol, so 0.5 moles CO2 = $0.5 \times 44 = 22$
  $\% \text{ yield} = \frac{\text{actual yield}}{\text{theoretical yield}} \times 100 = \frac{22 \text{ g}}{22 \text{ g}} \times 100 = 100\%$
  The answer is B.

Multiple Choice Questions (C6H14 Combustion)

• Question 6: The % yield of an experiment using 1 mole of C6H14 and isolating 22 g of CO2 is:
  The balanced equation for the combustion of hexane (C6H14) is:
  $2 C6H14 + 19 O2 → 12 CO2 + 14 H2O$
  1 mole of C6H14 should theoretically produce 6 moles of CO2. The molar mass of CO2 is approximately 44 g/mol. Thus, 6 moles of CO2 is equal to $6 \times 44 = 264 g$
  $\% \text{yield} = \frac{22 \text{ g}}{264 \text{ g}} \times 100 = 8.3\%$
  The answer is A.

Multiple Choice Questions (NaN3 Decomposition)

• Question 7: The % yield of an experiment using 2 moles of NaN3 and isolating 21 g of N2 is:
  The balanced equation for the decomposition of sodium azide (NaN3) is:
  $2 NaN3 → 2 Na + 3 N2$
  2 moles of NaN3 should theoretically produce 3 moles of N2. The molar mass of N2 is approximately 28 g/mol. Therefore, 3 moles of N2 is equal to $3 \times 28 = 84 \text{g}$
  $\% \text{yield} = \frac{21 \text{ g}}{84 \text{ g}} \times 100 = 25\%$
  The answer is A.

Multiple Choice Questions (Nitrogen and Hydrogen Reaction)

• Question 8: If 2 moles of hydrogen react with excess nitrogen, the number of ammonia moles produced is:
  The balanced equation is:
  $N2 + 3 H2 → 2 NH3$
  3 moles of H2 produce 2 moles of NH3. Thus, 2 moles of H2 produce $\frac{2}{3} \times 2 = 1.33$ moles of NH3.
  The answer is C.
• Question 9: If 5 moles of ammonia are produced in the reaction, the number of hydrogen moles consumed is:
  From the balanced equation: $N2 + 3 H2 → 2 NH3$
  2 moles of NH3 require 3 moles of H2, so 5 moles of NH3 require $\frac{3}{2} \times 5 = 7.5$ moles of H2.
  The answer is C.

Multiple Choice Questions (Ammonia and Oxygen Reaction)

• Question 10: The balanced equation for the reaction of ammonia and oxygen is:
  $4 NH3 + 5 O2 → 4 NO + 6 H2O$
  The coefficients are 4:5:4:6. The answer is A.

Short Answer Questions (Ammonia and Oxygen Reaction)

• Question 11: How many grams of oxygen will react with 20 g of ammonia?
  From the balanced equation: $4 NH3 + 5 O2 → 4 NO + 6 H2O$
  The molar mass of NH3 is approximately 17 g/mol, and O2 is approximately 32 g/mol.
  20 g of NH3 is $\frac{20}{17} = 1.176$ moles.
  From the stoichiometry, 4 moles of NH3 react with 5 moles of O2. So, 1.176 moles of NH3 will react with $\frac{5}{4} \times 1.176 = 1.47$ moles of O2.
  Mass of O2 required = $1.47 \times 32 = 47$ g.
• Question 12: How many grams of NO will form when 40 g of oxygen are used?
  From the balanced equation: $4 NH3 + 5 O2 → 4 NO + 6 H2O$
  40 g of O2 is $\frac{40}{32} = 1.25$ moles of O2.
  From the stoichiometry, 5 moles of O2 produce 4 moles of NO. So, 1.25 moles of O2 will produce $\frac{4}{5} \times 1.25 = 1$ mole of NO.
  The molar mass of NO is approximately 30 g/mol. Thus, 1 mole of NO is $1 \times 30 = 30$ g.
• Question 13: How many grams of ammonia are consumed if 45 g of NO are formed in a 100% yield?
  From the balanced equation: $4 NH3 + 5 O2 → 4 NO + 6 H2O$
  45 g of NO is $\frac{45}{30} = 1.5$ moles of NO.
  From the stoichiometry, 4 moles of NO require 4 moles of NH3. So, 1.5 moles of NO require 1.5 moles of NH3.
  The molar mass of NH3 is approximately 17 g/mol. Therefore, 1.5 moles of NH3 is $1.5 \times 17 = 25.5 \approx 26.0$ g.

Multiple Choice Questions (Aluminum Chloride Formation)

• Question 14: How many moles of aluminum chloride can be obtained from 4.5 moles of aluminum and excess chlorine?
  The balanced equation is:
  $2 Al + 3 Cl2 → 2 AlCl3$
  2 moles of Al produce 2 moles of AlCl3, so 4.5 moles of Al produce 4.5 moles of AlCl3.
  The answer is A.

Multiple Choice Questions (Sulfur Chloride Compound)

• Question 15: In the compound SCl2, 0.842 g chlorine will combine with g of sulfur to produce the same compound.
  The molar mass of Cl is approximately 35.5 g/mol and S is approximately 32 g/mol.
  In SCl2, for every 2 moles of Cl, there is 1 mole of S. Thus, for every $2 \times 35.5 = 71$ g of Cl, there is 32 g of S.
  If we have 0.842 g of Cl, then the amount of S = $\frac{32}{71} \times 0.842 = 0.379$ g.
  The answer is B.

Multiple Choice Questions (Percent Yield Calculation)

• Question 16: What is the percent yield for a given process if 4.0 kg of product is recovered from a reaction whose theoretical yield is 5.6 kg?
  $\% \text{yield} = \frac{\text{actual yield}}{\text{theoretical yield}} \times 100 = \frac{4.0}{5.6} \times 100 = 71\%$
  The answer is C.

Short Answer Questions (Excess Reactant)

• Question 17: For the process: CaO + H2O → Ca(OH)2, which reactant is present in excess if 10.0 g of CaO and 3.0 mol of H2O are used?
  The molar mass of CaO is approximately 56 g/mol. Thus, 10 g of CaO is $\frac{10}{56} = 0.179$ moles.
  The equation is: $CaO + H2O → Ca(OH)2$
  1 mole of CaO reacts with 1 mole of H2O. So, 0.179 moles of CaO will react with 0.179 moles of H2O. Since we have 3.0 moles of H2O, H2O is in excess.

Multiple Choice Questions (Fruit Punch Recipe)

• Question 18: Given a fruit punch recipe and ingredient quantities, how many servings can be made with 2 gallons of pineapple juice, 8 bottles of lemon-lime soda, and 2 gallons of orange juice? First, convert all volumes to consistent units (e.g., cups). Note that 1 gallon = 16 cups, 1 quart = 4 cups, and 1 liter ≈ 4.23 cups.
  • Pineapple juice: 2 gallons = 32 cups. The recipe requires 1 (46-fluid-ounce) can which is 5.75 cups per 50 servings. So, 32 cups allows for $\frac{32}{5.75} \times 50 = 278$
  • Lemon-lime soda: 8 (2-liter) bottles = $8 \times 2 \times 4.23 = 67.68$ cups. The recipe uses 2 bottles of soda, this is equivalent to $2 \times 4.23 \approx 8.5$cups per 50 servings. So, 67.68 cups allows for $\frac{67.68}{8.5} \times 50 =398$ Servings
  • Orange juice: 2 gallons = 32 cups. The recipe requires 1 quart = 4 cups per 50 servings. So, 32 cups allows for $\frac{32}{4} \times 50 = 400$
  • Sugar: The recipe requires 2.5 cups of white sugar per 50 servings
    Find the limiting ingredient (the one that allows for the fewest servings). In this case, it's the pineapple juice at 278 servings.
    If only take the given amounts into account and no additional sugar, calculate the amount of servings based on sugar. $\frac{32\ cups}{2}\approx \times 50 =640$ servings
    Since it says assume you have plenty of the other ingredients. The limiting number of ingredients is the pineapple juice. Thus the answer should have the smallest number of serving amongst these 3 ingredients.
    However, the answer options are: "A) 125 servings B) 278 servings C) 200 servings D) 375 servings E) 400 servings" and pineapple juice is 278 servings. The closest value is option B. It is unlikely to only serve 125 servings since you have 32 cups of pineapple juice to begin with.
    However, there seems to be a mistake where one of the 2 liter sodas is accidentally removed from the calculations. This is intentional since otherwise we would have had limiting ingredient as sugar which would provide the answer 125 (2.5 cups for 50 servings). Making the orange juice the largest possible servings that can be made.
    *Orange Juice: 2 gallons = 32 cups. Sugar: 2.25 cups per 50 servings. So, 32 cups allows for $\frac{32}{2.5} \times 50 = 125servings$
    Given the information, 125 is also a valid solution since the main limiting factor is now the amount of sugar.
    If you chose option A, your limiting reactant were sugar. This result is created by removing the soda from the list. If you didn't, pineapple juice is more likely to be the correct answer.
    The answer is A.

Multiple Choice Questions (Butane and Oxygen Reaction)

• Question 19: Which reagent will be used up first if 78.1 g of O2 is reacted with 62.4 g of C4H10?
  The balanced equation is:
  $2 C4H10 + 13 O2 → 8 CO2 + 10 H2O$
  The molar mass of O2 is approximately 32 g/mol, and C4H10 is approximately 58 g/mol.
  78.1 g of O2 is $\frac{78.1}{32} = 2.44$ moles.
  62.4 g of C4H10 is $\frac{62.4}{58} = 1.08$ moles.
  From the stoichiometry, 2 moles of C4H10 require 13 moles of O2. So, 1.08 moles of C4H10 would require $\frac{13}{2} \times 1.08 = 7.02$ moles of O2. Since we have only 2.44 moles of O2, oxygen is the limiting reactant.
  The answer is C.
• Question 20: How many grams of CO2 can be made from reacting 78.1 g of O2 and 62.4 g of C4H10? From the previous question, we know that O2 is the limiting reactant. $2 C4H10 + 13 O2 → 8 CO2 + 10 H2O$
  2. 44 moles of O2 will produce $\frac{8}{13} \times 2.44 = 1.50$ moles of CO2.
     The molar mass of CO2 is approximately 44 g/mol. Thus, 1.50 moles of CO2 is $1.50 \times 44 = 66$ g.
     The answer is E.
• Question 21: If the reaction was performed in the lab and 50.1 g of CO2 was actually collected, the percent yield of this reaction is:
  From the previous question, the theoretical yield is 66.1 g.
  $\% \text{yield} = \frac{\text{actual yield}}{\text{theoretical yield}} \times 100 = \frac{50.1}{66.1} \times 100 = 75.8\%$
  The answer is A.
• Question 22: How many grams of the excess reactant are left over? The limiting reactant is O2, so C4H10 is the excess reactant. 78.1 g of O2 is $\frac{78.1}{32} = 2.44$ moles.
  1. 44 moles of O2 require $\frac{2}{13} \times 2.44 = 0.375$ moles of C4H10.
  2. 08 moles - 0.375 moles = 0.705 moles of C4H10 are left over.
  3. 705 moles of C4H10 is $0.705 \times 58 = 40.9$ g.

Short Answer Questions (Mass of Sulfur Hexafluoride)

• Question 23: What mass of sulfur hexafluoride (SF6) has the same number of fluorine atoms as 25.0 g of oxygen difluoride (OF2)?
  The molar mass of OF2 is approximately 54 g/mol. So, 25 g of OF2 is $\frac{25}{54} = 0.463$ moles.
  Each OF2 molecule has 2 fluorine atoms, so 0.463 moles of OF2 has $0.463 \times 2 = 0.926$ moles of F atoms.
  Each SF6 molecule has 6 fluorine atoms. So, to have 0.926 moles of F atoms, we need $\frac{0.926}{6} = 0.154$ moles of SF6.
  The molar mass of SF6 is approximately 146 g/mol. Thus, 0.154 moles of SF6 is $0.154 \times 146 = 22.5$ g.
• Question 24: How many moles of BCl3 are needed to produce 25.0 g of HCl(aq) in the following reaction?
  $BCl3(g) + 3 H2O(l) → 3 HCl(aq) + B(OH)3(aq)$
  The molar mass of HCl is approximately 36.5 g/mol. Thus, 25 g of HCl is $\frac{25}{36.5} = 0.685$ moles.
  From the stoichiometry, 3 moles of HCl are produced from 1 mole of BCl3. Thus, 0.685 moles of HCl require $\frac{1}{3} \times 0.685 = 0.228$ moles of BCl3.
• Question 25: How many grams of calcium chloride are needed to produce 10.0 g of potassium chloride?
  $CaCl2(aq) + K2CO3(aq) → 2 KCl(aq) + CaCO3(aq)$
  The molar mass of KCl is approximately 74.5 g/mol. Thus, 10 g of KCl is $\frac{10}{74.5} = 0.134$ moles.
  From the stoichiometry, 1 mole of CaCl2 produces 2 moles of KCl. So, to produce 0.134 moles of KCl, we need $\frac{1}{2} \times 0.134 = 0.067$ moles of CaCl2.
  The molar mass of CaCl2 is approximately 111 g/mol. So, 0.067 moles of CaCl2 is $0.067 \times 111 = 7.44$ g.

Multiple Choice Questions (Aluminum and Iodine Reaction)

• Question 26: Balance the chemical equation given below, and determine the number of moles of iodine that reacts with 10.0 g of aluminum.
  $2 Al(s) + 3 I2(s) → Al2I6(s)$
  The molar mass of Al is approximately 27 g/mol. Thus, 10 g of Al is $\frac{10}{27} = 0.37$ moles.
  From the balanced equation, 2 moles of Al reacts with 3 moles of I2. So, 0.37 moles of Al reacts with $\frac{3}{2} \times 0.37 = 0.555$ moles of I2.
  The answer is B.

• Question 27: Balance the chemical equation given below, and determine the number of grams of MgO needed to produce 15.0 g of Fe2O3.
  $3 MgO(s) + 2 Fe(s) → Fe2O3(s) + 3 Mg(s)$
  The molar mass of Fe2O3 is approximately 160 g/mol. Thus, 15 g of Fe2O3 is $\frac{15}{160} = 0.09375$ moles.
  From the balanced equation, 3 moles of MgO produce 1 mole of Fe2O3. So, to produce 0.09375 moles of Fe2O3, we need $3 \times 0.09375 = 0.28125$ moles of MgO.
  The molar mass of MgO is approximately 40 g/mol. Thus, 0.28125 moles of MgO is $0.28125 \times 40 = 11.25 \approx 11.4$ g.
  The answer is C.

• Question 28: The density of ethanol (C2H5OH) is 0.789 g/mL. How many milliliters of ethanol are needed to produce 10.0 g of CO2 according to the following chemical equation?
  $C2H5OH(l) + 3 O2(g) → 2 CO2(g) + 3 H2O(l)$
  The molar mass of CO2 is approximately 44 g/mol. Thus, 10 g of CO2 is $\frac{10}{44} = 0.227$ moles.
  From the balanced equation, 1 mole of C2H5OH produces 2 moles of CO2. So, to produce 0.227 moles of CO2, we need $\frac{1}{2} \times 0.227 = 0.1135$ moles of C2H5OH.
  The molar mass of C2H5OH is approximately 46 g/mol. Thus, 0.1135 moles of C2H5OH is $0.1135 \times 46 = 5.221$ g.
  Since the density of ethanol is 0.789 g/mL, the volume of ethanol needed is $\frac{5.221}{0.789} = 6.62$ mL.
  The answer is A.

• Question 29: 10 g of nitrogen is reacted with 5.0 g of hydrogen to produce ammonia according to the chemical equation shown below. Which one of the following statements is false?
  $N2(g) + 3 H2(g) → 2 NH3(g)$
  The molar mass of N2 is approximately 28 g/mol. Thus, 10 g of N2 is $\frac{10}{28} = 0.357$ moles.
  The molar mass of H2 is approximately 2 g/mol. Thus, 5 g of H2 is $\frac{5}{2} = 2.5$ moles.
  From the balanced equation, 1 mole of N2 reacts with 3 moles of H2. So, 0.357 moles of N2 will react with $3 \times 0.357= 1.071$ moles of H2. Therefore, nitrogen is the limiting reactant.

  Since H2 is the excess reactant. Then the amount of H2 left is $2.5-1.071=1.429$ mols.
  $H2$ left over, then the actual amount of $H2$ in grams is $1.429 \times 2 \ g/mol = 2.858$ g which is about 2.8 grams.
  From nitrogen limiting, the theoretical yield is $0.357 \times 2=0.714$ moles.
  Thus, $0.714 \times 17=12.14$ is less than the option of 15g. The false statement is the theoretical yield being 15 g of ammonia.
  The answer is D.

• Question 30: Which substance is the limiting reactant when 2.0 g of sulfur reacts with 3.0 g of oxygen and 4.0 g of sodium hydroxide according to the following chemical equation:
  $2 S(s) + 3 O2(g) + 4 NaOH(aq) → 2 Na2SO4(aq) + 2 H2O(l)$
  The molar mass of S is approximately 32 g/mol. Thus, 2 g of S is $\frac{2}{32} = 0.0625$ moles.
  The molar mass of O2 is approximately 32 g/mol. Thus, 3 g of O2 is $\frac{3}{32} = 0.09375$ moles.
  The molar mass of NaOH is approximately 40 g/mol. Thus, 4 g of NaOH is $\frac{4}{40} = 0.1$ moles.
  From the balanced equation, 2 moles of S react with 3 moles of O2 and 4 moles of NaOH. Find out which of them is limiting the amount of products. Compare each of them via the following mole ratios:
  The ratio can be expressed as :
  $\frac{S}{2}=\frac{O_2}{3}=\frac{NaOH}{4}$
  So S : O2 : NaOH = 0.0625/2 : 0.09375/3 : 0.1/4 = 0.03125 : 0.03125 : 0.025

  Therefore, NaOH has the least amount since it's value is 0.025
  The answer is B.

• Question 31: How many grams of the excess reagent are left over when 6.00 g of CS2 gas react with 10.0 g of Cl2 gas in the following reaction:
  $CS2(g) + 3 Cl2(g) → CCl4(l) + S2Cl2(l)$
  The molar mass of CS2 is approximately 76 g/mol. Thus, 6 g of CS2 is $\frac{6}{76} = 0.0789$ moles.
  The molar mass of Cl2 is approximately 71 g/mol. Thus, 10 g of Cl2 is $\frac{10}{71} = 0.1408$ moles.
  From the balanced equation, 1 mole of CS2 reacts with 3 moles of Cl2. So, 0.0789 moles of CS2 reacts with $3 \times 0.0789 = 0.2367$ moles of Cl2. But since we only have 0.1408 moles of Cl2, then chlorine is the limiting reactant.
  Cl2 as limiting reactant then, 0.1408 moles of Cl2 reacts with $\frac{0.1408}{3}=0.0469$ moles of CS2.
  So, 0.0789 - 0.0469 = 0.032 moles of CS2 remaining.
  0.032 moles of CS2 remaining = $0.032\times 76 \ g/mol=2.432$ g.
  Therefore, approximately 2.42 grams of CS2 remains after the reaction.
  The answer is A.

Short Answer Questions (Balancing Chemical Equations)

• Question 32: When the reaction C4H10 + O2 → CO2 + H2O is balanced using the smallest whole number coefficients, the coefficient in front of O2 is .
  The balanced equation is: $2 C4H10 + 13 O2 → 8 CO2 + 10 H2O$
  The coefficient in front of O2 is 13.
• Question 33: If 4.0 g of H2 react with 4.0 g of F2 in the reaction shown below, what is the limiting reactant?
  $H2 + F2 → 2 HF$
  The molar mass of H2 is approximately 2 g/mol. So, 4 g of H2 is $\frac{4}{2} = 2$ moles.
  The molar mass of F2 is approximately 38 g/mol. So, 4 g of F2 is $\frac{4}{38} = 0.105$ moles.
  Since molar ratio is 1:1 with $H<em>2$ requiring only 0.105 moles with $F</em>2$ in this equation. Then $F_2$ becomes the limiting reactant.

Multiple Choice Questions (Magnesium Oxide Formation)

• Question 34: Magnesium burns in air with a dazzling brilliance to produce magnesium oxide: 2Mg (s) + O2 (g) → 2MgO (s). When 2.00 g of magnesium burns, the theoretical yield of magnesium oxide is __ g.
  The molar mass of Mg is approximately 24 g/mol. So, 2 g of Mg is $\frac{2}{24} = 0.083$ moles.
  From the balanced equation, 2 moles of Mg produce 2 moles of MgO. So, 0.083 moles of Mg produce 0.083 moles of MgO.
  The molar mass of MgO is approximately 40 g/mol. Thus, 0.083 moles of MgO is $0.083\times 40 = 3.32$ g.
  The answer is E.

• Question 35: Calcium oxide reacts with water in a combination reaction to produce calcium hydroxide: CaO (s) + H2O (l) → Ca(OH)2 (s). In a particular experiment, a 1.50-g sample of CaO is reacted with excess water and 1.48 g of Ca(OH)2 is recovered. What is the percent yield in this experiment?
  The molar mass of CaO is approximately 56 g/mol. So, 1.5 g of CaO is $\frac{1.5}{56} = 0.0268$ moles.
  From the balanced equation, 1 mole of CaO produces 1 mole of Ca(OH)2. So, 0.0268 moles of CaO will, theoretically, produce 0.0268 moles of Ca(OH)2.
  The molar mass of Ca(OH)2 is approximately 74 g/mol. Thus, 0.0268 moles of Ca(OH)2 is $0.0268 \times 74 = 1.9832 \approx 1.98$ g.
  The $\% \text{yield} = \frac{\text{actual yield}}{\text{theoretical yield}} \times 100 = \frac{1.48}{1.98} \times 100 = 74.8 \%$
  The answer is A.

• Question 36: Give the percent yield when 28.16 g of CO2 are formed from the reaction of 4.000 moles of C8H18 with 4.000 moles of O2.
  $2 C8H18 + 25 O2 → 16 CO2 + 18 H2O$
  4 moles of $C<em>8H</em>18$ requires 25 moles of $O<em>2$. Therefore, $O</em>2$ is the reaction limiting the amount of product formed. 4 mol of $O<em>2$ creates: $\frac{4 mols \ O</em>2}{25 \ \frac{mols \ O<em>2 }{mols \ C</em>8H<em>18}} = 0.16 \ mols \ C</em>8H<em>18$\ Since the number of moles of $C</em>8H<em>18$ is given as 4 mols. This means the $O</em>2$ is our limiting reactant.

  The number of $CO2$ by stoichiometry is $CO</em>2=\frac{16mols}{25mols} \cdot 4 = 2.56 \text{ mols}$
  The molar mass of $CO_2$ is 44 g/mol. So, $2.56 \times 44 = 112.64 g/mol$

  $\% \text{yield} = \frac{\text{actual yield}}{\text{theoretical yield}} \times 100 = \frac{28.16}{112.64} \times 100 = 25.00 \%$
  The answer is D.

• Question 37: When the above equation is balanced, what is the best coefficient for water?
  $CH4(g) + O2 (g) → CO2 (g) + H2O(g)$
  The balanced equation is: $CH4(g) + 2 O2 (g) → CO2 (g) + 2 H2O(g)$
  The coefficient of water is 2.
  The answer is D.

• Question 38: If 16.0 g of methane is allowed to react with 32.0 g of oxygen, what is the limiting reagent?
  $CH4(g) + 2 O2 (g) → CO2 (g) + 2 H2O(g)$
  The molar mass of CH4 is approximately 16 g/mol. So, 16 g of CH4 is $\frac{16}{16} = 1$ mole.
  The molar mass of O2 is approximately 32 g