Question

Aqueous sulfuric acid will react with solid sodium hydroxide to produce aqueous sodium sulfate and liquid water . Suppose 8.8 g of sulfuric acid is mixed with 6.48 g of sodium hydroxide. Calculate the maximum mass of water that could be produced by the chemical reaction. Be sure your answer has the correct number of significant digits.

Answer 1

Answer on Question #48206 - Chemistry - Other

Question

Aqueous sulfuric acid will react with solid sodium hydroxide to produce aqueous sodium sulfate and liquid water. Suppose 8.8 g of sulfuric acid is mixed with 6.48 g of sodium hydroxide. Calculate the maximum mass of water that could be produced by the chemical reaction. Be sure your answer has the correct number of significant digits.

**Answer**: Balanced reaction equation is:

$$\mathrm{H_2SO_4(aq)} + 2\mathrm{NaOH(s)} = \mathrm{Na_2SO_4(aq)} + 2\mathrm{H_2O(l)}$$

Molar masses of the reactants equal:

$$\mathrm{M(H_2SO_4)} = 98\ \mathrm{g/mol},\quad \mathrm{M(NaOH)} = 40\ \mathrm{g/mol}$$

Number of moles of the reactants are:

$$n(H_2SO_4) = \frac{m(H_2SO_4)}{M(H_2SO_4)} = \frac{8.8}{98} = 0.090\ \mathrm{mol}$$ $$n(NaOH) = \frac{m(NaOH)}{M(NaOH)} = \frac{6.48}{40} = 0.16\ \mathrm{mol}$$

Then we make a proportion:

- 1 mole of $\mathrm{H_2SO_4}$ reacts with 2 moles of NaOH

- 0.090 moles of $\mathrm{H_2SO_4} - x$ moles of NaOH

$$x = \frac{0.090 \cdot 2}{1} = 0.18\ \mathrm{moles}\ \text{of}\ \mathrm{NaOH}\ \text{should react with}\ 0.090\ \mathrm{moles}\ \text{of}\ \mathrm{H_2SO_4}$$

There are only 0.16 moles of sodium hydroxide, therefore it is the limiting reactant.

We need to make another proportion to calculate the maximum mass of $\mathrm{H_2O}$ that could be produced by the chemical reaction:

- 2 moles of NaOH produce 2 moles of $\mathrm{H_2O}$

- 0.16 moles of $\mathrm{NaOH} - x$ moles of $\mathrm{H_2O}$

$$x = \frac{0.16 \cdot 2}{2} = 0.16\ \mathrm{moles}\ \text{of}\ \mathrm{H_2O}\ \text{could be produced}$$

The mass of $\mathrm{H_2O}$ equals:

$$m(H_2O) = n(H_2O) \cdot M(H_2O) = 0.16 \cdot 18 = 2.9\ \mathrm{g}$$

**Answer**: 2.9 g of $\mathrm{H_2O}$

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