1. Write the balanced chemical equation for the reaction:
H2SO4 + 2NaOH -> Na2SO4 + 2H2O
2. Determine the limiting reactant:
- Convert the mass of each reactant to moles:
52.0 g H2SO4 x (1 mol H2SO4 / 98.08 g H2SO4) = 0.530 mol H2SO4
19.9 g NaOH x (1 mol NaOH / 40.00 g NaOH) = 0.498 mol NaOH
- Calculate the stoichiometry for each reactant:
For H2SO4: 0.530 mol H2SO4 x (2 mol H2O / 1 mol H2SO4) = 1.06 mol H2O
For NaOH: 0.498 mol NaOH x (2 mol H2O / 2 mol NaOH) = 0.498 mol H2O
The limiting reactant is NaOH because it produces less water.
3. Calculate the theoretical yield of water:
0.498 mol H2O x (18.02 g H2O / 1 mol H2O) = 8.98 g H2O
Round to 3 significant figures:
The theoretical yield of water formed from the reaction is 9.00 g.
Aqueous sulfuric acid (H2SO4) reacts with solid sodium hydroxide (NaOH) to produce aqueous sodium sulfate (Na2SO4) and liquid water (H20). What is the theoretical yield of water formed from the reaction of 52.0 g of sulfuric acid and 19.9 g of sodium hydroxide?
Round your answer to 3 significant figures.
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