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 75.5g of sulfuric acid and 105.g of sodium hydroxide

109.34 g

Explanation:

2NaOH(aq) + H2SO4(aq) > Na2SO4(aq) + 2H2O(l)

Number of moles of NaOH = 105g/40g/mol = 2.6 moles

From the reaction equation;

2 moles of NaOH yields 1 mole of sodium sulphate

2.6 moles of NaOH yields = 2.6 × 1/2 = 1.3 moles of sodium sulphate

Number of moles of H2SO4= 75.5g/98 g/mol = 0.77 moles

From the reaction equation;

1 mole of H2SO4 yields 1 mole of sodium sulphate

Hence, 0.77 moles of H2SO4 yields 0.77 moles of sodium sulphate

So H2SO4 is the limiting reactant.

Theoretical yield = number of moles × molar mass

= 0.77 mol ×142 g/mol

= 109.34 g

i believe the answer is three water molecules