Question

If 0.3 mole Zn are added to HCl containing 0.52 mole of HCl, how many moles of $H_{2}$ are produced, according to the reaction

$Zn\left(\right.s\left.\right)+2HCl\left(\right.a\left.\right) \rightarrow ZnCl_{2}\left(\right.a\left.\right)+H_{2}\left(\right.g\left.\right)$

• A. 1 mole
• B. 2 mole
• C. 0.8 mole
• D. 0.26 mole

Answer 1

Answer: (D)

$Zn\left(\right.s\left.\right)+2HCl\left(\right.a\left.\right) \rightarrow ZnCl_{2}\left(\right.a\left.\right)+H_{2}\left(\right.g\left.\right)$

As we have 0.30 mole of Zn, therefore, HCl will react completely, i.e., HCl is the limiting reactant.

2 moles of HCl produce $H_{2}=1$ mole

So 0.52 mole of HCl will produce

$H_{2}=\frac{1}{2}\times 0.52$ mole

$=0.26$ mole