Step 1
The reaction must be written, completed, and balanced:
2 NaOH + H2SO4 => Na2SO4 + H2O
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Step 2
Information provided:
Mass of NaOH = 40.0 g
Mass of H2SO4 = 60.0 g
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Information needed:
The molar masses:
NaOH) 40.0 g/mol approx.
H2SO4) 98.0 g/mol approx.
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Step 3
The limiting reactant?
By stoichiometry:
1 mole NaOH = 40.0 g
1 mole H2SO4 = 98.0 g
2 NaOH + H2SO4 => Na2SO4 + H2O
2 x 40.0 g NaOH ----------- 98.0 g H2SO4
40.0 g NaOH ----------- X
X = 40.0 g NaOH x 98.0 g H2SO4/2 x 40.0 g NaOH = 49.0 g H2SO4
For 40.0 g of NaOH, 49.0 g of H2SO4 is needed but is provided 60.0 g of H2SO4. Therefore, the excess is the H2SO4, and the limiting reactant is the NaOH.
Answer: the limiting reactant is NaOH