Question

Q.6. 1.2 g of Mg was dropped in 100 mL of 0.1 M HCl solution.Calculate the volume of H2 obtained by STP.

Answer 1

Equation 
Mg + 2HCl → MgCl2 + H2 
1mol Mg react with 2mol HCl to produce 1 mol H2 
Check for limiting reactant : 
Molar mass Mg = 24g/mol 
mol in 1.2g Mg = 1.2/24 = 0.05 mol Mg 
This will react with 0.05*2 = 0.10 mol HCl 
Mol HCl in 100mL of 0.1M HCl solution = 100/1000*0.1 = 0.01 mol HCl 
The HCl is the limiting reactant 

From the equation: 
0.01 mol HCl will produce 0.01/2 = 0.005 mol H2 
At STP , 1 mol H2 gas has volume of 22.4L 
At STP , 0.005 mol H2 gas has volume = 0.005*22.4 = 0.112 L H2gas. 

The volume of H2 produced is 0.112L