Question

Find the amount of zinc in gram necessary to produce $224ml$ of $H_2$ by the reaction with sulphuric acid.

Answer 1

Given, Reaction between $Zn$ and Acid to produce $H_2$ is

$Zn+H_{2}S{O}_4⟶ZnS{O}_4+H_2$

From above reaction, it is clear that $1$ mole of $Zn$ produces $1$ moles of $H_2$

$\therefore$ $224$ ml of $Zn$ was required to produce $224$ ml $H_2$.

Now we know

$1$ mole=$22.4L$ of $Zn$

$\Rightarrow 224ml$ of $Zn=224\times {10}^{-3}L$ of $Zn$

$\Rightarrow 224\times {10}^{-3}L$ of $Zn=\frac{224\times {10}^{-3}}{22.4}$ moles of $Zn$

$\therefore$ No. of moles required=${10}^{-2}$ moles of $Zn$

We know, $1$ mole of $Zn=65gms$ of $Zn$

$\Rightarrow {10}^{-2}$ moles of $Zn=65\times {10}^{-2}gms$ of $Zn$

$\therefore$ Amount of Zinc in gms necessary to produce $224ml$ of $H_2=0.65gms$.