Question

If $49$ grams of $H_{2}S{O}_4$ react with $80.0$ grams of NaOH, how much reactant will be left over once the reaction is complete?

• A. $24.5$g $H_{2}S{O}_4$
• B. None of either compound
• C. $20.$g NaOH
• D. $40.$g NaOH
• E. $60.$g NaOH

Answer 1

Answer: (D)

$40.$g NaOH

$H_{2}S{O}_4+2NaOH\to N{a}_{2}S{O}_4+2H_{2}O$
The molar masses of $H_{2}S{O}_4$ and NaOH are 98 g/mol and 40 g/mol respectively.
Number of moles is the ratio of mass to molar mass.
Number of moles of $H_{2}S{O}_4=\frac{49}{98}=0.5$ moles
Number of moles of NaOH $=\frac{80.0}{40}=2$ moles
0.5 moles of $H_{2}S{O}_4$ will react with $2\times 0.5=1.0$ mole of NaOH.
$2-1=1$ mole of NaOH will remain unreacted. This corresponds to $1\times 40=40g$.