# 1.0 g of magnesium is burnt with 0.56 g O2 in a closed vessel. Which reactant is left in excess and how much?(At. wt.Mg = 24; O = 16) Option 1) Mg, 0.16 g Option 2) O2 , 0.16 g Option 3) Mg, 0.44 g Option 4) O2, 0.28 g

The reaction is as follows

$M_{g}+\frac{1}{2}O_{2}\rightarrow M_{g}O$

modes of $M_{g}=\frac{1}{24}=0.04167$

modes of $O_{2}=\frac{0.56}{32}=0.0175$

According to reaction stoichiometry $\frac{1}{24 }$ modes of $M_{g}$ require $\frac{1}{48}$ modes of $O_{2}$ or 0.02083 modes of $O_{2}$

clearly $O_{2}$ is the limiting reagent modes of $M_{g}$ required by 0.01175 modes of $O_{2}=0.0175\times 2=0.035ml$

$\therefore$ Excess $M_{g}=0.04167-0.035$

$=0.00667ml$

$=0.16 g$

Option 1)

Mg, 0.16 g

Correct Option

Option 2)

O2 , 0.16 g

Incorrect Option

Option 3)

Mg, 0.44 g

Incorrect Option

Option 4)

O2, 0.28 g

Incorrect Option

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