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

 

Answers (1)

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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