Question

0.04 g of magnesium was put into the bottom of the tube containing a dilute solution of $HCl$ and it reacted completely and formed hydrogen gas and the following data was recorded.
Air pressure in the room = 730 mm Hg
Temperature of the water solution = 302 K
Vapor pressure of water at 302 K = 30 mm Hg
The gas collected did not fill the eudiometer. The height of the meniscus above the level of the water was 40.8 mm.
For the complete reaction of 0.04 g of Mg theoretical yield (in mL) at STP of hydrogen gas is :

• A. 10 mL
• B. 25 mL
• C. 37 mL
• D. 46 mL
• E. 51 mL

Answer 1

Answer: (C)

37 mL

Molar mass of Mg is 24.1 g/mol.
0.040 g of mg corresponds to $\frac{0.040}{24.1}=0.00166$ moles.
$Mg+2HCl\to MgC{l}_2+H_2$
1 mole of Mg gives 1 mole of hydrogen.
0.00166 moles of Mg will give 0.00166 moles of hydrogen.
At STP, 1 mole of hydrogen will occupy a volume of 22.4 L.
0.00166 moles of hydrogen will occupy a volume of $22.4\times 0.00166=0.037L=37mL$.
This is theoretical yield.