Question

2Mg(s) + O${}_2$(g) $\to$ 2MgO(s)
If 48.6 grams of magnesium are placed in a container with 64 grams of oxygen gas and the reaction above proceeds to completion, what is the mass of MgO(s) produced?

• A. $15.4$ grams
• B. $32.0$ grams
• C. $80.6$ grams
• D. $96.3$ grams
• E. $112$ grams

Answer 1

The correct option is C $80.6$ grams

$2Mg+O_2\to 2MgO$
The atomic masses of Mg and O are 24.3 g/mol and 16 g/mol respectively.
The molar mass of $MgO=24.3+16=40.3g/mol$
Number of moles is the ratio of mass to molar/atomic mass.
48.6 g of Mg corresponds to $\frac{48.6}{24.3}=2moles$
64 g of O2 corresponds to $\frac{64}{2\times 16}=2moles$
2 moles of Mg will react with 1 mole of $O_2$ to produce 2 moles of $MgO$.
Since 2 moles of $O_2$ are present, $O_2$ is excess reagent and $Mg$ is limiting reagent.
2 moles of Mg will give 2 moles of $MgO$.
Mass of MgO $=2\times 40.3=80.6g$