Question

How much reactant remains if $92$ g $HN{O}_3$ reacts with $24$ g $LiOH$ assuming a complete reaction?

• A. $46$ g $HN{O}_3$
• B. $29$ g $HN{O}_3$
• C. $12$ g $HN{O}_3$
• D. $2$ g $LiOH$
• E. $12$ g $LiOH$

Answer 1

The correct option is B $29$ g $HN{O}_3$

The balanced chemical equation is
$HN{O}_3+LiOH\to LiN{O}_3+H_{2}O$
Thus, 1 mole of $HN{O}_3$ reacts with 1 mole of $LiOH$.
The molar masses of $HN{O}_3$ and $LiOH$ are 63 g/mol and 24 g/mol respectively.
92 g $HN{O}_3$ corresponds to $\frac{92}{63}=1.46$ moles.
24 g $LiOH$ corresponds to $\frac{24}{24}=1$ mole.
Thus 1 mole of $LiOH$ will react with 1 mole of $HN{O}_3$ but 1.46 moles of $HN{O}_3$ are present.
Hence, $1.46-1=0.46$ moles of $HN{O}_3$ will remain unreacted.
The mass of $HN{O}_3$ remaining will be $0.46\times 63=29$ g.