Question

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 water formed from the reaction of 4.9g of sulfuric acid and 3.7g of sodium hydroxide? Be sure your answer has the correct number of significant digits in it.

Answer 1

molar mass of H2SO4 = 98 g/mol
molar mass of NaOH = 40 g/mol
number of mole = (given mass)/(molar mass)
number of mole of H2SO4 = 4.9/98
= 0.05 mole
number of mole of NaOH = 3.7/40
= 0.092 mole

reaction taking place is
H2SO4 + 2NaOH --> Na2SO4 + 2H2O
according to reaction
1 mol of H2SO4 required 2 mol of NaOH
0.05 mol of H2SO4 required (2*0.05) mol of NaOH
0.05 mol of H2SO4 required 0.10 mol of NaOH

but we have only 0.092 mole of NaOH
so, NaOH is limiting reagent

again,
2 mol of NaOH give 2 mol of water
0.092 mol of NaOH give 0.092 mol of water

so,
number of mol of water formed = 0.092 mole
molar mass of water = 18 g/mol
mass of water formed = (number of mol of water formed)*(molar mass of water)
= 0.092*18
= 1.7 g

Answer : 1.7 g