Question

Aqueous sulfuric acid will react with solid sodium hydroxide to produce aqueous sodium sulfate and liquid water . Suppose 0.98 g of sulfuric acid is mixed with 0.240 g of sodium hydroxide. Calculate the maximum mass of sodium sulfate that could be produced by the chemical reaction. Be sure your answer has the correct number of significant digits.

Answer 1

Answer:

0.43 grams is the maximum mass of sodium sulfate that could be produced by the chemical reaction.

Explanation:

Mass of sulfuric acid = 0.98 g

Moles of sulfuric acid = $\frac{0.98 g}{98 g/mol}=0.010 mol$

Mass of sodium hydroxide = 0.240 g

Moles of sodium hydroxide = $\frac{0.240 g}{40 g/mol}=0.0060 mol$

$H_2SO_4(aq)+2NaOH(s)\rightarrow Na_2SO_4(aq)+2H_2O(l)$

According to reaction, 2 moles of sodium hydroxide reacts with 1 mole of sulfuric acid , then 0.0060 moles of sodium hydroxide will react with :

$\frac{1}{2}\times 0.0060 mol=0.0030 mol$ of sulfuric acid

As we can see that we have 0.010 moles of sulfuric acid but only 0.0030 moles of sulfuric acid will react which indicates that it is in excessive amount where as sodium hydroxide is in limiting amount.

So, amount of sodium sulfate to be formed will depend upon moles of sodium hydroxide.

According to reaction, 2 moles of sodium hydroxide gives with 1 mole of sodium sulfate , then 0.0060 moles of sodium hydroxide will give :

$\frac{1}{2}\times 0.0060 mol=0.0030 mol$ of sodium sulfate

Mass of 0.0030 moles of sodium sulfate :

0.0030 mol × 142 g/mol = 0.426 g ≈ 0.43 g

0.43 grams is the maximum mass of sodium sulfate that could be produced by the chemical reaction.