gen chem- Stoichiometry

A solution contains 85 g of NaCl dissolved in water, resulting in a total volume of 500 mL. If the density of the solution is 1.12 g/mL, what is the mass percent (m/m%) of NaCl in the solution?

A. 85 / (85+100)(1.12)(100)

B. 85 / (500)(1.12)

C. (85)(100) / (85+500)(1.12)

D. (85)(100) / (85+100)

E. (85)(100) / (500)(1.12)

E

• mass percent = mass of solute x 100 / mass of solution

• mass of solution = total volume x density of solution

  • 500 × 1.12

The mass percentage of oxygen in an oxygen-nitrogen compound is known. All of the following are needed to determine the molecular formula of the compound EXCEPT one. Which is the EXCEPTION?

A. empirical formula

B. molar mass of the compound

C. atomic mass of nitrogen

D. atomic mass of oxygen

E. Avogadro’s number

E

A 2 L solution is prepared by dissolving 30 g of a compound with a molar mass of 120 g/mol in water. What is the molarity of the solution?

A. (30)(2) / 120

B. (30)(12) / 2

C. 30 / (120)(2)

D. 2 / (30)(120)

E. (120)(2) / 30

C

Molarity (M) = moles solute / solution volume

to solve for the moles of solute

• molar mass = compound mass / moles compound

  120 g/mol = 30g / moles compound

  moles compound = 30 / 120

The following reaction proceeds with a percent yield of 50%. Calculate the actual amount of NH₃ created when 28 g of N₂ reacts with excess H₂ according to the following balanced chemical equation: 

3H₂ + N₂ → 2NH₃

A. 17g

B. 22g

C. 28g

D. 32g

A

• find theoretical yield of NH3

  28g N2 x (1molN2/28gN2) x (2molNH3/1molN2) x (17gNH3/1molNH3) = 34g NH3

• percent yield = actual yield / theoretical yield

  0.5 = actual / 34

  actual = 0.5 × 34 = 17

Given the reaction shown below, if 3.0 L of 2.0 M KOH reacts completely with excess H₂SO₄, how many moles of K₂SO₄ are produced?

H₂SO₄ + 2KOH → K₂SO₄ + 2H₂O

A. 2.0

B. 3.0

C. 4.0

D. 6.0

E. 12.0

B

• to determine the amnt of K2SO4 produced, calculate the # of moles of KOH in the rxn

  # of mol KOH = 2.0M x 3.0L = 6 mol

• calculate the # of moles of K2SDO4

  6molKOH x (1molK2SO4/2molsKOH) = 3 mols K2SO4

How many moles of water can be produced from 11.2 liters of hydrogen gas and excess oxygen gas at STP?

2H₂ + O₂ → 2H₂O

A. 0.25

B. 0.50

C. 1.00

D. 2.00

B

• at STP, 1 mole of any gas has a volume of 22.4 L

• calculate the moles of H2

  11.2 L H2 / 22.4 L/mol = 0.5 mol H2

• solve for moles of water produced

  0.5molH2 x (2molH20/2molH2) = 0.5mol H2O

The Haber process, a chemical reaction shown below, is run to completion, producing 216 g of NH₃. However, the calculated yield of the experiment is 1.3 kg. What is the percent yield of the reaction?

N₂ + 3H₂ → 2NH₃

A. 0.216 / 1.3

B. (216)(100) / (1.3)

C. (0.216)(100) / (1.3)

D. (1300) / (216)(100)

E. (1300)(100) / (216)

C

• percent yield = actual x 100 / theoretical

• the actual yield is 216g but we need to convert it to kg b/c the theoretical yield is 1.3kg

• so 216 / 1000 = 0.216 = actual yield

34 grams of NH₃ (17 g/mol) reacted to completion according to the following balanced chemical equation:

2NH₃ + CO₂ → CN₂OH₄ + H₂O

If 20 grams of CN₂OH₄ (60 g/mol) were obtained, what is the percent yield?

A. 25%

B. 33%

C. 50%

D. 66%

E. 75%

B

34gNH3 x (1molNH3/17gNH3) x (1molCN2OH4/2molNH3) x (60g/1mol) = 60g

percent yield = 20/60 = 1/3 = 33%

How many H atoms are present in 0.001 mol of CH₄?

A. 2.41 × 10²¹

B. 2.41 × 10²²

C. 6.02 × 10²⁰

D. 6.02 × 10²¹

E. 6.02 × 10²²

A

0.001molCH4 x (6.02 × 10²³ / 1 mol) = 6.02 × 10²⁰ molecules CH4

6.02 × 10²⁰ molecules CH4 x (4 H atoms / 1 CH4 molecule) = 2.41 × 10²¹ H atoms

A reaction between A and B to form C involving 1 mole of A and 1 mole of B is allowed to proceed. Experimental data shows that in the reaction, 0.6 moles of B reacts, 0.2 moles of A remains unreacted, and 0.2 moles of C is produced. 

What is the balanced chemical equation for this reaction?

A. 2A + B → C

B. 2A + 4B → C

C. 4A + 2B → C

D. 4A + 3B → 2C

E. 4A + 3B → C

E

• 0.6 moles of compound B is involved in the rxn

• 0.8 moles of compound A is involved in the rxn

• 0.2 moles of compound C is created in the rxn

0.8A + 0.6B → 0.2C

• multiply coefficients by a common number so there are no decimals

0.8×5= 4

0.6×5= 3

0.2×5= 1

Which of the following would be the empirical formula of a compound that has 6 moles of carbon and 15 moles of hydrogen?

A. CH4

B. C2H4

C. C2H5

D. C3H6

E. C3H8

C

• to find the empirical formula of the given compound, calculate the mole ratio b/w the carbon and hydrogen by dividing by the smaller mole #

  6/6 moles of carbon = 1 mole carbon

  15/6 moles of hydrogen = 2.5 mole hydrogen

• multiply 2.5 by 2 to get rid of decimal

  1×2= 2 mole carbon

  2.5×2= 5 mole hydrogen