StoiChiometry

Here are practice questions for each topic covered in the images, formatted as Question followed by Answer below.

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Topic: Writing & Balancing Equations

Question 1:

Balance the following equation:

\text{CH}_4 + \text{O}_2 \rightarrow \text{CO}_2 + \text{H}_2\text{O}

Answer:

\text{CH}_4 + 2\text{O}_2 \rightarrow \text{CO}_2 + 2\text{H}_2\text{O}

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Question 2:

Balance the following equation:

\text{NaOH} + \text{H}_2\text{SO}_4 \rightarrow \text{Na}_2\text{SO}_4 + \text{H}_2\text{O}

Answer:

2\text{NaOH} + \text{H}_2\text{SO}_4 \rightarrow \text{Na}_2\text{SO}_4 + 2\text{H}_2\text{O}

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Topic: Types of Chemical Reactions

Question 3:

Identify the type of reaction:

2\text{Mg} + \text{O}_2 \rightarrow 2\text{MgO}

Answer:

Synthesis reaction

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Question 4:

Identify the type of reaction:

\text{NaOH} + \text{HCl} \rightarrow \text{H}_2\text{O} + \text{NaCl}

Answer:

Acid/Base reaction (neutralization)

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Topic: Prediction of Products

Question 5:

Predict the products of the single replacement reaction:

\text{Al} + \text{CuCl}_2 \rightarrow

Answer:

2\text{Al} + 3\text{CuCl}_2 \rightarrow 3\text{Cu} + 2\text{AlCl}_3

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Question 6:

Predict the products of the complete combustion of propane:

\text{C}_3\text{H}_8 + \text{O}_2 \rightarrow

Answer:

\text{C}_3\text{H}_8 + 5\text{O}_2 \rightarrow 3\text{CO}_2 + 4\text{H}_2\text{O}

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Topic: Mole to Mole Stoichiometry

Question 7:

Using the equation \text{N}_2 + 3\text{H}_2 \rightarrow 2\text{NH}_3, how many moles of NH_3 can be made from 4.0 moles of H_2?

Answer:

4.0 \, \text{mol H}_2 \times \frac{2 \, \text{mol NH}_3}{3 \, \text{mol H}_2} = 2.67 \, \text{mol NH}_3

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Question 8:

How many moles of CuCl_2 are needed to completely react with 0.53 mol Al in the reaction:

2\text{Al} + 3\text{CuCl}_2 \rightarrow 2\text{AlCl}_3 + 3\text{Cu}

Answer:

0.53 \, \text{mol Al} \times \frac{3 \, \text{mol CuCl}_2}{2 \, \text{mol Al}} = 0.795 \, \text{mol CuCl}_2

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Topic: Mass to Mass Stoichiometry

Question 9:

What mass of chlorine gas (Cl_2) is needed to react completely with 20.0 g of sodium?

2\text{Na} + \text{Cl}_2 \rightarrow 2\text{NaCl}

(Atomic masses: Na = 23.0 g/mol, Cl = 35.5 g/mol)

Answer:

20.0 \, \text{g Na} \times \frac{1 \, \text{mol Na}}{23.0 \, \text{g}} \times \frac{1 \, \text{mol Cl}_2}{2 \, \text{mol Na}} \times \frac{71.0 \, \text{g}}{1 \, \text{mol Cl}_2} = 30.9 \, \text{g Cl}_2

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Topic: Volume & Molecules

Question 10:

How many molecules of BaSO_4 are produced from 18.2 g of Ba(OH)_2?

\text{Ba(OH)}_2 + \text{H}_2\text{SO}_4 \rightarrow \text{BaSO}_4 + 2\text{H}_2\text{O}

(Molar mass Ba(OH)_2 = 171.3 g/mol)

Answer:

18.2 \, \text{g Ba(OH)}_2 \times \frac{1 \, \text{mol}}{171.3 \, \text{g}} \times \frac{1 \, \text{mol BaSO}_4}{1 \, \text{mol Ba(OH)}_2} \times 6.02 \times 10^{23} = 6.40 \times 10^{22} \, \text{molecules}

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Topic: Energy in Reactions

Question 11:

How much energy is released when 25 g of NO_2 reacts?

2\text{NO}_2 \rightarrow \text{N}_2\text{O}_4 + 56 \, \text{kJ}

(Molar mass NO_2 = 46.0 g/mol)

Answer:

25 \, \text{g NO}_2 \times \frac{1 \, \text{mol}}{46.0 \, \text{g}} \times \frac{56 \, \text{kJ}}{2 \, \text{mol NO}_2} = 15.2 \, \text{kJ}

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Topic: Percent Yield

Question 12:

When 36.4 g of CaCO_3 reacts with excess HCl, 38.2 g of CaCl_2 is produced.

\text{CaCO}_3 + 2\text{HCl} \rightarrow \text{CaCl}_2 + \text{H}_2\text{CO}_3

Calculate the percent yield.

(Molar masses: CaCO_3 = 100.1 g/mol, CaCl_2 = 111.0 g/mol)

Answer:

Theoretical yield:

36.4 \, \text{g CaCO}_3 \times \frac{1 \, \text{mol}}{100.1 \, \text{g}} \times \frac{1 \, \text{mol CaCl}_2}{1 \, \text{mol CaCO}_3} \times \frac{111.0 \, \text{g}}{1 \, \text{mol}} = 40.4 \, \text{g CaCl}_2

Percent yield:

\frac{38.2}{40.4} \times 100\% = 94.6\%