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.
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
Get Answers For Free
Most questions answered within 1 hours.