The amount of Zinc (atomic weight =65 ) necessary to produce 224 mL of H 2 by the reaction with an acid will be

Question

The amount of Zinc (atomic weight =65 ) necessary to produce 224 mL of H 2 by the reaction with an acid will be (A) 0.65 g (B) 7.6 g (C) 6.5 g (D) 8.5 g

Tardigrade Admin · Accepted Answer

Atomic weight of Zn =95 and volume of H 2=224 mL Zn + H 2 SO 4 longrightarrow ZnSO 4+ H 2 ( text Volume of H 2/22400)=( text Weight of Zn / text Atomic weight of Zn ) Wt of Z n=(224 × 65/22400)=0.65 g

Q. The amount of Zinc (atomic weight $= 65$ ) necessary to produce $224 m L$ of $H_{2}$ by the reaction with an acid will be

0.65 g

7.6 g

6.5 g

8.5 g

Atomic weight of $Z n = 95$ and volume of $H_{2} = 224 m L$
$Z n + H_{2} S O_{4} ⟶ Z n S O_{4} + H_{2}$
$\frac{Volume of H _{2}}{22400} = \frac{Weight of Z n}{Atomic weight of Z n}$
Wt of $Z n = \frac{224 \times 65}{22400} = 0.65 g$

Q. The amount of Zinc (atomic weight =65 ) necessary to produce 224mL of H2​ by the reaction with an acid will be