If 0.3 mole Zn are added to HCl containing 0.52 mole of HCl, how many moles of H2 are produced, according to the reaction Zn(.s.)+2HCl(.a.) arrow ZnCl2(.a.)+H2(.g.)

Question

If 0.3 mole Zn are added to HCl containing 0.52 mole of HCl, how many moles of H2 are produced, according to the reaction Zn(.s.)+2HCl(.a.) arrow ZnCl2(.a.)+H2(.g.) (A) 1 mole (B) 2 mole (C) 0.8 mole (D) 0.26 mole

Tardigrade Admin · Accepted Answer

Zn(.s.)+2HCl(.a.) arrow ZnCl2(.a.)+H2(.g.) 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 H2=1 mole So 0.52 mole of HCl will produce H2=(1/2)× 0.52 mole =0.26 mole

Q. 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

$Z n (s) + 2 H Cl (a) \to Z n C l_{2} (a) + H_{2} (g)$

1 mole

2 mole

0.8 mole

0.26 mole

$Z n (s) + 2 H Cl (a) \to Z n C l_{2} (a) + H_{2} (g)$

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

Q. If 0.3 mole Zn are added to HCl containing 0.52 mole of HCl, how many moles of H2​ are produced, according to the reaction Zn(s)+2HCl(a)→ZnCl2​(a)+H2​(g)