Question

Aqueous sulfuric acid

H2SO4

will react with solid sodium hydroxide

NaOH

to produce aqueous sodium sulfate

Na2SO4

and liquid water

H2O

. Suppose 71. g of sulfuric acid is mixed with 22.5 g of sodium hydroxide. Calculate the maximum mass of water that could be produced by the chemical reaction. Be sure your answer has the correct number of significant digits.

Answer 1

Balanced equation is

H2SO4 + 2NaOH ------> Na2SO4 + 2H2O

number of moles of sulfuric acid = 71g / 98.079 g/mol = 0.724 mole

number of moles of NaOH = 22.5g / 40.0 g/mol = 0.5625 mole

from the balanced equation we can say that

1 mole of H2So4 requires 2 mole of NaOH so

0.724 mole of H2SO4 will require

= 0.724 mole of H2SO4 *(2 mole of NaOH / 1 mole of H2So4)

= 1.448 mole of NaOH

but we have only 0.5625 mole of NaOH which is the limiting reactant

from the balanced equation we can say that

2 mole of NaOH produces 2mole of H2O so

0.5625 mole of NaOH will produce 0.5625 mole of H2O

1 mole of H2O = 18.016g

0.5625 mole of H2O = 10.1 g

Therefore, the mass of H2O produced will be 10.1 g