Question

what volume of 0.311 m sulfurous acid solution could be neutralized by 2.00 g aluminum hydroxide? express you answer in milliliters.

Explanation:

Step1: Write balanced reaction

$$\ce{2Al(OH)3 + 3H2SO3 -> Al2(SO3)3 + 6H2O}$$

Step2: Calculate moles of $\ce{Al(OH)3}$

Molar mass of $\ce{Al(OH)3} = 27 + 3\times(16+1) = 78\ \text{g/mol}$
$$n_{\ce{Al(OH)3}} = \frac{2.00\ \text{g}}{78\ \text{g/mol}} \approx 0.02564\ \text{mol}$$

Step3: Relate moles to $\ce{H2SO3}$

From reaction: $\frac{n_{\ce{H2SO3}}}{n_{\ce{Al(OH)3}}} = \frac{3}{2}$
$$n_{\ce{H2SO3}} = 0.02564\ \text{mol} \times \frac{3}{2} = 0.03846\ \text{mol}$$

Step4: Calculate volume of acid

Molarity $M = \frac{n}{V}$, so $V = \frac{n}{M}$
$$V = \frac{0.03846\ \text{mol}}{0.311\ \text{mol/L}} \approx 0.1237\ \text{L}$$

Step5: Convert to milliliters

$$V = 0.1237\ \text{L} \times 1000 = 123.7\ \text{mL}$$

Answer:

124 mL (rounded to 3 significant figures)