Questions: The chemical equation shows how ammonia reacts with sulfuric acid to produce ammonium sulfate. 2 NH3(aq) + H2SO4(aq) → (NH4)2SO4(aq) How many grams of ammonium sulfate can be produced if 60.0 mol of sulfuric acid react with an excess of ammonia 1,020 g 3,970 g 5,890 g 7,930 g

Solution Steps

The balanced chemical equation is:

\[ 2 \mathrm{NH}_{3}(\mathrm{aq}) + \mathrm{H}_{2} \mathrm{SO}_{4}(\mathrm{aq}) \rightarrow \left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4}(\mathrm{aq}) \]

This equation tells us that 1 mole of sulfuric acid (\(\mathrm{H}_{2} \mathrm{SO}_{4}\)) reacts with 2 moles of ammonia (\(\mathrm{NH}_{3}\)) to produce 1 mole of ammonium sulfate (\(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4}\)).

Given that there is an excess of ammonia, sulfuric acid is the limiting reactant. We have 60.0 moles of sulfuric acid, and according to the stoichiometry of the reaction, 1 mole of sulfuric acid produces 1 mole of ammonium sulfate.

Therefore, 60.0 moles of sulfuric acid will produce 60.0 moles of ammonium sulfate.

The molar mass of ammonium sulfate (\(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4}\)) is calculated as follows:

• Nitrogen (N): \(14.01 \, \text{g/mol} \times 2 = 28.02 \, \text{g/mol}\)
• Hydrogen (H): \(1.008 \, \text{g/mol} \times 8 = 8.064 \, \text{g/mol}\)
• Sulfur (S): \(32.07 \, \text{g/mol} \times 1 = 32.07 \, \text{g/mol}\)
• Oxygen (O): \(16.00 \, \text{g/mol} \times 4 = 64.00 \, \text{g/mol}\)

Adding these together gives the molar mass of ammonium sulfate:

\[ 28.02 + 8.064 + 32.07 + 64.00 = 132.134 \, \text{g/mol} \]

Now, convert moles to grams:

\[ 60.0 \, \text{moles} \times 132.134 \, \text{g/mol} = 7,928.04 \, \text{g} \]

Final Answer