Question

3 g of Mg is burnt in a closed vessel containing 3 g of oxygen. If the percentage yield of the reaction is 80%, the weight of MgO produced is:
$2Mg+O_2\to 2MgO$

• A. 5 g
• B. 4 g
• C. 10 g
• D. 8 g

Answer 1

The correct option is B 4 g

Molar mass of Mg = 24 g/mol
Number of moles of Mg $=\frac{3}{24}mol$
$=0.125mol$
Molar mass of $O_2=32$ g/mol
Number of moles of $O_2=\frac{3}{32}mol$
$=0.094mol$
The given reaction is:
$2Mg+O_2\to 2MgO$

$\frac{\text{moles of Mg}}{\text{stoichiometric coefficient}}=\frac{0.125}{2}=0.0625$

$\frac{\text{moles of oxygen}}{\text{stoichiometric coefficient}}=\frac{0.094}{1}=0.094$

So, $O_2$ is the excess reagent here.

2 moles of Mg produce $2\times 0.8$ moles of MgO (since yield is 80% given)

Hence, 0.125 moles of Mg will produce
$=\frac{2\times 0.8}{2}\times 0.125$ mol MgO
= 0.1 mol MgO
Molar mass of MgO = 40 g

$\therefore$ Mass of MgO produced $=40\times 0.1g$
= 4 g