Stoichiometry Practice Problems – Step-by-Step Guidance

Q1a. How many moles and what mass of chlorine gas (Cl2) are required to react with 10.0 g of sodium metal (Na) to produce sodium chloride (NaCl)?

Background

Topic: Stoichiometry (Chemical Reactions and Moles)

This question tests your ability to use stoichiometry to relate the mass of a reactant (sodium) to the amount (in moles and grams) of another reactant (chlorine gas) needed for a reaction.

Key Terms and Formulas

• Mole: A unit that measures the amount of substance (1 mole = 6.022 x 1023 particles).

• Molar Mass: The mass of one mole of a substance (g/mol).

• Stoichiometry: The calculation of reactants and products in chemical reactions.

Balanced equation for the reaction:

$2\,\mathrm{Na} + \mathrm{Cl}_2 \rightarrow 2\,\mathrm{NaCl}$

Step-by-Step Guidance

1. Calculate the number of moles of sodium ($\mathrm{Na}$) you have using its molar mass ($22.99\,\mathrm{g/mol}$):

   $\text{moles Na} = \frac{10.0\,\mathrm{g}}{22.99\,\mathrm{g/mol}}$

2. Use the stoichiometric coefficients from the balanced equation to find the mole ratio between $\mathrm{Na}$ and $\mathrm{Cl}_2$:

   $2\,\mathrm{Na} : 1\,\mathrm{Cl}_2$

3. Set up a proportion to convert moles of $\mathrm{Na}$ to moles of $\mathrm{Cl}_2$ needed:

   $\text{moles Cl}_2 = \text{moles Na} \times \frac{1\,\mathrm{mol}\,\mathrm{Cl}_2}{2\,\mathrm{mol}\,\mathrm{Na}}$

4. Calculate the mass of $\mathrm{Cl}_2$ required using its molar mass ($70.90\,\mathrm{g/mol}$):

   $\text{mass Cl}_2 = \text{moles Cl}_2 \times 70.90\,\mathrm{g/mol}$

Try solving on your own before revealing the answer!

Final Answer:

Number of moles of $\mathrm{Cl}_2$ required: 0.217 mol

Mass of $\mathrm{Cl}_2$ required: 15.4 g

We first calculated moles of sodium, used the stoichiometric ratio to find moles of chlorine gas, and then converted that to grams using the molar mass of $\mathrm{Cl}_2$.

Q1b. How many moles and what mass (in g) of sodium nitrate (NaNO3) are required to decompose and produce 128 g of diatomic oxygen (O2), where NaNO2 is the other product?

Background

Topic: Stoichiometry (Decomposition Reactions)

This question tests your ability to use stoichiometry to relate the mass of a product (oxygen gas) to the amount (in moles and grams) of a reactant (sodium nitrate) required for a decomposition reaction.

Key Terms and Formulas

• Decomposition Reaction: A single compound breaks down into two or more products.

• Molar Mass of $\mathrm{O}_2$: $32.00\,\mathrm{g/mol}$

• Molar Mass of $\mathrm{NaNO}_3$: $85.00\,\mathrm{g/mol}$

Balanced equation for the decomposition:

$2\,\mathrm{NaNO}_3 \rightarrow 2\,\mathrm{NaNO}_2 + \mathrm{O}_2$

Step-by-Step Guidance

1. Calculate the number of moles of $\mathrm{O}_2$ produced from the given mass:

   $\text{moles O}_2 = \frac{128\,\mathrm{g}}{32.00\,\mathrm{g/mol}}$

2. Use the stoichiometric coefficients from the balanced equation to find the mole ratio between $\mathrm{O}_2$ and $\mathrm{NaNO}_3$:

   $2\,\mathrm{NaNO}_3 : 1\,\mathrm{O}_2$

3. Set up a proportion to convert moles of $\mathrm{O}_2$ to moles of $\mathrm{NaNO}_3$ required:

   $\text{moles NaNO}_3 = \text{moles O}_2 \times \frac{2\,\mathrm{mol}\,\mathrm{NaNO}_3}{1\,\mathrm{mol}\,\mathrm{O}_2}$

4. Calculate the mass of $\mathrm{NaNO}_3$ required using its molar mass:

   $\text{mass NaNO}_3 = \text{moles NaNO}_3 \times 85.00\,\mathrm{g/mol}$

Try solving on your own before revealing the answer!

Final Answer:

Number of moles of $\mathrm{NaNO}_3$ required: 8.00 mol

Mass of $\mathrm{NaNO}_3$ required: 680 g

We calculated moles of $\mathrm{O}_2$, used the stoichiometric ratio to find moles of $\mathrm{NaNO}_3$, and then converted to grams.

Q1c. How many moles of carbon dioxide and molecules of water are formed by the combustion of 20.0 kg of glucose in excess oxygen?

Background

Topic: Stoichiometry (Combustion Reactions)

This question tests your ability to use stoichiometry to relate the mass of a reactant (glucose) to the amount (in moles and molecules) of products (CO2 and H2O) formed in a combustion reaction.

Key Terms and Formulas

• Combustion Reaction: A substance reacts with oxygen to produce CO2 and H2O.

• Molar Mass of Glucose ($\mathrm{C}_6\mathrm{H}_{12}\mathrm{O}_6$): $180.16\,\mathrm{g/mol}$

• Avogadro's Number: $6.022 \times 10^{23}$ molecules/mol

Balanced equation for glucose combustion:

$\mathrm{C}_6\mathrm{H}_{12}\mathrm{O}_6 + 6\,\mathrm{O}_2 \rightarrow 6\,\mathrm{CO}_2 + 6\,\mathrm{H}_2\mathrm{O}$

Step-by-Step Guidance

1. Convert the mass of glucose from kg to g:

   $20.0\,\mathrm{kg} \times 1000 = 20,000\,\mathrm{g}$

2. Calculate the number of moles of glucose:

   $\text{moles glucose} = \frac{20,000\,\mathrm{g}}{180.16\,\mathrm{g/mol}}$

3. Use the stoichiometric coefficients to find moles of CO2 and H2O produced per mole of glucose:

   $1\,\mathrm{mol}\,\mathrm{C}_6\mathrm{H}_{12}\mathrm{O}_6$ produces $6\,\mathrm{mol}\,\mathrm{CO}_2$ and $6\,\mathrm{mol}\,\mathrm{H}_2\mathrm{O}$

4. Calculate the number of molecules of water using Avogadro's number:

   $\text{molecules H}_2\mathrm{O} = \text{moles H}_2\mathrm{O} \times 6.022 \times 10^{23}$

Try solving on your own before revealing the answer!

Final Answer:

Moles of CO2 formed: 666 mol

Molecules of H2O formed: $2.41 \times 10^{27}$ molecules

We converted mass to moles, used stoichiometry to find moles of products, and then used Avogadro's number for molecules.

Q1d. Calculate the mass of magnesium oxide produced if 2.40 g of magnesium reacts with 10.0 g of oxygen gas.

Background

Topic: Limiting Reactant and Stoichiometry

This question tests your ability to identify the limiting reactant and use stoichiometry to calculate the mass of product formed.

Key Terms and Formulas

• Limiting Reactant: The reactant that is completely consumed first, limiting the amount of product formed.

• Molar Mass of Mg: $24.31\,\mathrm{g/mol}$

• Molar Mass of O2: $32.00\,\mathrm{g/mol}$

• Molar Mass of MgO: $40.31\,\mathrm{g/mol}$

Balanced equation:

$2\,\mathrm{Mg} + \mathrm{O}_2 \rightarrow 2\,\mathrm{MgO}$

Step-by-Step Guidance

1. Calculate the number of moles of Mg and O2 you have:

   $\text{moles Mg} = \frac{2.40\,\mathrm{g}}{24.31\,\mathrm{g/mol}}$

   $\text{moles O}_2 = \frac{10.0\,\mathrm{g}}{32.00\,\mathrm{g/mol}}$

2. Use the stoichiometric coefficients to determine which reactant is limiting:

   $2\,\mathrm{Mg} : 1\,\mathrm{O}_2$

3. Calculate the maximum moles of MgO that can be formed based on the limiting reactant:

   $2\,\mathrm{Mg} \rightarrow 2\,\mathrm{MgO}$ (1:1 ratio)

4. Calculate the mass of MgO produced using its molar mass:

   $\text{mass MgO} = \text{moles MgO} \times 40.31\,\mathrm{g/mol}$

Try solving on your own before revealing the answer!

Final Answer:

Mass of MgO produced: 3.98 g

We determined the limiting reactant, calculated moles of product, and converted to grams using the molar mass of MgO.