Question

3g of Mg is burnt in a closed vessel containing 3 g of oxygen. The weight of excess reactant left is
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

Mg + 1/2O2--> MgO
24g 16g
Here limiting is Mg as it is required more.
1gm Mg requires 16/24g Oxygen
3gm Mg requires 3×2/3=2 gm of oxygen.
so 3 gm Mg and 2 gm O is reacted.

what left is 1 gm of Oxygen.

Answer 2

Answer : The weight of excess reactant left is, (B) 1.0 g of oxygen.

Solution : Given,

Mass of Mg = 3 g

Mass of $O_2$ = 3 g

Molar mass of Mg = 24 g/mole

Molar mass of $O_2$ = 32 g/mole

First we have to calculate the moles of $Mg$ and $O_2$

$\text{ Moles of Mg}=\frac{\text{ Mass of Mg}}{\text{ Molar mass of Mg}}=\frac{3g}{24g/mole}=0.125moles$

$\text{ Moles of }O_2=\frac{\text{ Mass of }O_2}{\text{ Molar mass of }O_2}=\frac{3g}{32g/mole}=0.094moles$

• The balanced chemical reaction is,

$2Mg+O_2\rightarrow 2MgO$

From the reaction, we conclude that

2 moles of $Mg$ react with 1 moles of $O_2$

0.125 moles of $Mg$ react with $\frac{0.125}{2}=0.0625$ moles of $O_2$

So, oxygen is present in excess amount and magnesium in less amount.

In this reaction, $Mg$ is limiting reactant and $O_2$ is excess reactant.

Moles of excess reactant left = 0.094 - 0.0625 = 0.0315 moles

Weight of excess reactant left = Moles of excess reactant left × Molar mass of excess reactant

Weight of excess reactant left = 0.0315 g × 32 g/mole = 1.008 g = 1.0 g

Therefore, the weight of excess reactant left is, (B) 1.0 g of oxygen.