Question

6.5g of Zn is dissolved in excess of H2SO4. The weight of ZnSO4 formed and the volume of H2 gas liberated at STP are (Zn=65)

Answer 1

this is your solution

Answer 2

Answer:

The weight of $ZnSO_{4}$ equals to g will be formed.

The volume of $H_{2}$  gas liberated at STP is equal to $2.24L$.

Explanation:

The balanced chemical equation:

      $Zn+H_{2}SO_{4}$   →   $ZnSO_{4}+H_{2}$

Given mass of Zn  $=6.5g$

Number of moles of Zn $=\frac{6.5}{65}=0.1moles$

From the equation: one mole of Zn produced $=1mole$ of H₂

Volume of 1mole of $H_{2}=22.4L$

Then, volume of 0.1mole of $H_{2}=(22.4)(0.1)=2.24L$

Again, from balanced equation:

1 mole of Zn produced$=1mole$ of $ZnSO_{4}$

0.1 mole Zn will produce $=0.1mole$ of $ZnSO_{4}$

We know, $Weight= (moles)(Molar Mass)$

Molar mass of $ZnSO_{4}=161gmol^{-1}$

$W_{ZnSO_{4}}=(0.1)(161)=16.1g$