Question

$100$ mL solution of $NaOH$ (containing $4$ g $NaOH$ per litre) and $50$ mL of $HCl$ (containing $7.3$ g $HCl$ per litre) react as follows,
$NaOH\left(aq\right)+HCl\left(aq\right)⟶NaCl\left(aq\right)+H_{2}O\left(l\right)$
$0.5$ g of $NaCl$ is formed. Thus, unreacted $NaOH$ is:

• A. $0.058$ g
• B. $3.66$ g
• C. $10.8$ g
• D. $0.63$ g

Answer 1

The correct option is A $0.058$ g

$NaOH\left(aq\right)+HCl\left(aq\right)⟶NaCl\left(aq\right)+H_{2}O\left(l\right)$
$0.4$ g $0.365$ g
$0.01$ mol $0.01$ mol
So, according to this, $0.585$ g of $NaCl$ should be formed but, only $0.5$ g forms.
Remaining $NaOH$ is $0.4\times \frac{(0.585-0.5)}{0.585}=0.058$ g.