Aqueous sulfuric acid (H2SO4) reacts with solid sodium hydroxide (NaOH) to produce aqueous sodium sulfate (Na2SO4) and liquid water (H2O). What is the theoretical yield of sodium sulfate formed from the reaction of 6.87 g of sulfuric acid and 10.0 g of sodium hydroxide?

Question

Be sure your answer has the correct number of significant digits in it.

Answer 1

To calculate the theoretical yield of sodium sulfate (\( \text{Na}_2\text{SO}_4 \)) from the reaction between sulfuric acid (\( \text{H}_2\text{SO}_4 \)) and sodium hydroxide (\( \text{NaOH} \)), we can start with the balanced chemical equation for the reaction:

\[ \text{H}_2\text{SO}_4 (aq) + 2 \text{NaOH} (s) \rightarrow \text{Na}_2\text{SO}_4 (aq) + 2 \text{H}_2\text{O} (l) \]

Step 1: Calculate the moles of each reactant.

1. Moles of sulfuric acid (\( \text{H}_2\text{SO}_4 \)):

   The molar mass of \( \text{H}_2\text{SO}_4 \) is calculated as: \[ 2(1.01) + 32.07 + 4(16.00) = 98.09 , \text{g/mol} \]

   Now, we find the moles of \( \text{H}_2\text{SO}_4 \): \[ \text{Moles of } \text{H}_2\text{SO}_4 = \frac{6.87 , \text{g}}{98.09 , \text{g/mol}} \approx 0.0700 , \text{mol} \]

2. Moles of sodium hydroxide (\( \text{NaOH} \)):

   The molar mass of \( \text{NaOH} \) is: \[ 22.99 + 1.01 + 16.00 = 40.00 , \text{g/mol} \]

   Now, we find the moles of \( \text{NaOH} \): \[ \text{Moles of } \text{NaOH} = \frac{10.0 , \text{g}}{40.00 , \text{g/mol}} = 0.250 , \text{mol} \]

Step 2: Determine the limiting reactant.

According to the balanced equation, 1 mole of \( \text{H}_2\text{SO}_4 \) reacts with 2 moles of \( \text{NaOH} \).

From the moles calculated:

• Moles of \( \text{H}_2\text{SO}_4 \) = 0.0700 mol
• To react completely with 0.0700 mol of \( \text{H}_2\text{SO}_4 \), we need: \[ 2 \times 0.0700 , \text{mol} = 0.140 , \text{mol of NaOH} \]

We have 0.250 mol of \( \text{NaOH} \), which is more than enough. Therefore, \( \text{H}_2\text{SO}_4 \) is the limiting reactant.

Step 3: Calculate the theoretical yield of sodium sulfate.

From the balanced equation, 1 mole of \( \text{H}_2\text{SO}_4 \) produces 1 mole of \( \text{Na}_2\text{SO}_4 \).

Thus, the moles of \( \text{Na}_2\text{SO}_4 \) produced will be equal to the moles of the limiting reactant \( \text{H}_2\text{SO}_4 \): \[ \text{Moles of } \text{Na}_2\text{SO}_4 = 0.0700 , \text{mol} \]

Now we calculate the mass of \( \text{Na}_2\text{SO}_4 \): The molar mass of \( \text{Na}_2\text{SO}_4 \) is: \[ 2(22.99) + 32.07 + 4(16.00) = 142.04 , \text{g/mol} \]

Now, we find the mass of \( \text{Na}_2\text{SO}_4 \): \[ \text{Mass of } \text{Na}_2\text{SO}_4 = 0.0700 , \text{mol} \times 142.04 , \text{g/mol} \approx 9.94 , \text{g} \]

Final Result:

The theoretical yield of sodium sulfate, rounded to the correct number of significant digits (three significant figures), is: \[ \boxed{9.94 , \text{g}} \]