How many moles of hydrogen will be evolved when 6.50 g of zinc (Atomic weight = 65) is dipped in  $6.023\times {10}^{22}$ molecules of hydrochloric acid at STP $(\text{Zn}+2\text{HC}l\to \text{ZnC}{l}_2+{\text{H}}_{\text{2}})$ $\left({\text{AS}}_{\text{1}}\right)$

The displacement reaction is

$\text{Zn}\left(s\right)+2\text{HC}l\left(aq\right)\to \text{\hspace{0.17em}\hspace{0.17em}ZnC}{l}_2\left(aq\right)+{\text{\hspace{0.17em}\hspace{0.17em}H}}_2\left(g\right)$

$65\text{\hspace{0.17em}g\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}2\times 6.023\times {10}^{23}\to 1 \text{\hspace{0.17em}mole}$

$6.5\text{\hspace{0.17em}g\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}6.023\times {10}^{22}\to x \text{\hspace{0.17em}mole}$

The stoichiometric relation between  Zn and  ${\text{H}}_2$ is

$\text{Zn}\to {\text{H}}_2$

$65\text{\hspace{0.17em}g}\to 1\text{\hspace{0.17em}mole}$

$6.5\text{\hspace{0.17em}g}\to x\text{\hspace{0.17em}mole}$

$\therefore x=0.1{\text{\hspace{0.17em}mole\hspace{0.17em}of\hspace{0.17em}H}}_2$

The relation between HCl and ${\mathbf{\text{H}}}_{\mathbf{2}}$ is

$2\text{\hspace{0.17em}HC}l\to {\text{H}}_2$

$2\times 6.023\times {10}^{23}\to 1\text{\hspace{0.17em}mole}$

$6.023\times {10}^{22}\to x\text{\hspace{0.17em}mole}$

$\therefore x=\frac{1.0{\text{\hspace{0.17em}mol\hspace{0.17em}H}}_{\text{2}}}{2\times 6.023\times {10}^{23}}\times 6.023\times {10}^{22}=0.05{\text{\hspace{0.17em}moles\hspace{0.17em}of\hspace{0.17em}H}}_2$

Here, the limiting reagent is HCl because small amount of product ${\text{H}}_2$ is formed.