Question

1 g of $Mg$ is burnt in a vessel containing 0.5 g of oxygen. The reactant remaining unreacted is:

• A. 0.24 g of $Mg$
• B. 0.1 g of $Mg$
• C. 0.1 g of $O_2$
• D. 0.75 g of $Mg$

Answer 1

The correct option is A 0.24 g of $Mg$

$2Mg+O_2\to 2MgO$

(1 g $Mg$) / (24.30506 g $Mg$/mol) = 0.0411 mol $Mg$

(0.5 g $O_2$) / (31.99886 g $O_2$/mol) = 0.0156 mol $O_2$

0.0156 mole of $O_2$ would react completely with 0.0156 $\times$ (2/1) = 0.0312 mole of $Mg$, but there is more $Mg$ present than that, so $Mg$ is in excess.

((0.0411 mol $Mg$ initially) - (0.0312 mol $Mg$ reacted)) $\times$ (24.30506 g $Mg$/mol) = 0.24 g $Mg$ left over