Question

How many grams of ${Na}_2{SO}_4$ can be produced by reacting $98$ g $H_2{SO}_4$ with $40$ g $NaOH$ ?

• A. $18$ g
• B. $36$ g
• C. $71$ g
• D. $142$ g
• E. $150$ g

Answer 1

The correct option is C $71$ g

The balanced chemical equation is
$H_{2}S{O}_4+2NaOH\to N{a}_{2}S{O}_4+2H_{2}O$
Thus, 1 mole $H_{2}S{O}_4$ will react with 2 moles of $NaOH$.
The molecular weights of $H_{2}S{O}_4$ and $NaOH$ are 98 g/mol and 40 g/mol respectively.
The mass is divided with molecular weight to obtain the number of moles.
The number of moles of $H_{2}S{O}_4=\frac{98}{98}=1$ mole.

The number of moles of $NaOH$ $=\frac{40}{40}=1$ mole.
Thus 1 mole of $H_{2}S{O}_4$ and 1 mole of $NaOH$ are present.
0.5 mole of $H_{2}S{O}_4$ will react with 1 mole of $NaOH$.

Thus $H_{2}S{O}_4$ is the excess reagent and $NaOH$ is limiting reactant.
1 mole of $H_{2}S{O}_4$ will produce 1 mole of $N{a}_{2}S{O}_4$
Hence, 0.5 mole of $H_{2}S{O}_4$ will produce 0.5 mole of $N{a}_{2}S{O}_4$
The molecular weight of $N{a}_{2}S{O}_4$ is 142 g/mol.
The mass of $N{a}_{2}S{O}_4$ produced is $0.5\times 142=71$ g.