Question

$1.0 \, g$ of Mg is burnt with 0.28 g of $O_2$ in a closed vessel. Which reactant is left in excess and how much ?

• A. Mg, 5.8 g
• B. Mg, 0.58 g
• C. $O_2 , 0.24 \, g$
• D. $O_2 , 2.4 \, g$

Answer 1

Answer: (B)

Mg, 0.58 g

Answer 2

$2Mg + O_2 \to 2MgO$

48 g of Mg requires 32 g of $O_2$ ?

requires 0.28 g of $O_2$

Mass of magnesium required = $\frac{48 \times 0.28}{32} = 0.42 \, g$
But 1 g of Mg is available. Thus, Mg is the excess reagent.
Excess of Mg left behind $= 1 - 0.42 = 0.58\, g$