How many moles of HCl are contained in 3 kg 5% chlorhydric acid solution ?
Let's refer to the $5\%$ chlorhydric acid solution by $S$:
$m_S=3000g$
$%_{HCl}=5$
1) $m_{HCl}$=$\frac{\%_{HCl}\cdot m_S}{100}$=$\frac{5 \cdot 3000}{100}$=$150g$
(don't get confused with $kg$ and $g$ !)
$M_{HCl}$=$1+35.5$=$36.5\frac{g}{mol}$
2) $n_{HCl}$=$\frac{m_{HCl}}{M_{HCl}}$=$ \frac{150}{36,5}$=$4.11 mol$
