Question

3 g of $Mg$ is burnt in a closed vessel containing 3 g of oxygen. The weight of excess reactant left is:
$2Mg+O_2\to 2MgO$

• A. 0.5 g of oxygen
• B. 1.0 g of oxygen
• C. 1.0 g of $Mg$
• D. 0.5 g of $Mg$

Answer 1

The correct option is B 1.0 g of oxygen

$2Mg+O_2\to 2MgO$
Moles of $Mg=\frac{3}{24}=0.125$ mol
Moles of oxygen $=\frac{3}{32}=0.094$ mol
Finding the limiting reagent
For $Mg=\frac{0.125}{2}=0.0625$
For $O_2=\frac{0.094}{1}=0.094$
So, Magnesium is the limiting reagent.
2 moles of magnesium reacts with 1 mole of oxygen
$(2\times 24)$ g of magnesium reacts with 32 g of oxygen
3 g of magnesium will react with $=\frac{32}{2\times 24}\times 3=2$ g of oxygen
Amount of excess reactant left = 1 g of oxygen