Question

1.0 g of magnesium is burnt with 0.56 g $O_2$ in a closed vessel. Which reactant is left in excess and how much ? (At.wt. Mg=24; O=16)

• A. Mg, 0.16 g
• B. $O_2$, 0.16 g
• C. Mg, 0.44 g
• D. $O_2$, 0.28 g

Answer 1

Answer: (A)

Mg, 0.16 g

The reaction involved is:

$2Mg+O_2⟶2MgO$

Now, No. of moles of $Mg$ = $\frac{1.0g}{2.4g/mol}=0.0416$ moles

No. of moles of $O_2$ = $\frac{0.56g}{32g/mol}=0.0175$ moles

$1$ mole $O_2$ consumes $2$ mole $Mg$ & gives $2$ mole $MgO$

$\Rightarrow O_2$ here is limiting reagent as $0.0416$ moles of $Mg$ require $0.0208$ mole of $O_2$ but only $0.0175$ moles of $O_2$ given.

$\Rightarrow 0.0175$ moles $O_2$ will consume $2\times 0.0175$ moles of $Mg$

$\therefore$ No. of moles of $Mg$ consumed = $0.035$ moles

$\Rightarrow$ Remaining moles of $Mg=0.0416-0.035$

$=0.0066$ moles of $Mg$

As we know, $1$ mole of $Mg=24g$

$\Rightarrow 0.0066$ moles of $Mg=24\times 0.006=0.1584g$

$\therefore$ Amount of $Mg$ left in excess = $0.16g$

Hence, option A is correct.