1.0 g of Mg is burnt with 0.28 g of O2 in a closed vessel. Which reactant is left in excess and how much ?

Question

1.0 g of Mg is burnt with 0.28 g of O2 in a closed vessel. Which reactant is left in excess and how much ? (A) Mg, 5.8 g (B) Mg, 0.58 g (C) O2 , 0.24 g  (D) O2 , 2.4 g

Tardigrade Admin · Accepted Answer

2Mg + O2 → 2MgO 48 g of Mg requires 32 g of O2 ? requires 0.28 g of O2 Mass of magnesium required = (48 × 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

Q. $1.0 g$ of Mg is burnt with 0.28 g of $O_{2}$ in a closed vessel. Which reactant is left in excess and how much ?

Mg, 5.8 g

Mg, 0.58 g

$O_{2}, 0.24 g$

$O_{2}, 2.4 g$

$2 M g + O_{2} \to 2 M g O$

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 $= 1 - 0.42 = 0.58 g$

Q. 1.0g of Mg is burnt with 0.28 g of O2​ in a closed vessel. Which reactant is left in excess and how much ?

Mg, 5.8 g

Mg, 0.58 g

O2​,0.24g

O2​,2.4g

2Mg+O2​→2MgO 48 g of Mg requires 32 g of O2​ ? requires 0.28 g of O2​ Mass of magnesium required = 3248×0.28​=0.42g But 1 g of Mg is available. Thus, Mg is the excess reagent. Excess of Mg left behind =1−0.42=0.58g

Q. $1.0 g$ of Mg is burnt with 0.28 g of $O_{2}$ in a closed vessel. Which reactant is left in excess and how much ?

$O_{2}, 0.24 g$

$O_{2}, 2.4 g$

$2 M g + O_{2} \to 2 M g O$

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 $= 1 - 0.42 = 0.58 g$