# Question #2c774

Dec 5, 2017

Roughly $2491 k g$

#### Explanation:

The reaction between $M g$ and ${O}_{2}$ to produce $M g O$ goes like this:

$2 M {g}_{\left(s\right)} + {O}_{2 \left(g\right)} \to 2 M g {O}_{\left(s\right)}$

$4131 k g = 4131000 g$

$n \left(M g O\right) = \frac{4131000}{24.3 + 16} = 102506.2035 m o l$

Molar ratio of Mg:MgO is 1:1

So, $102506.2035 m o l$ of $M g$ are needed.

${M}_{r} \left(M g\right) = 24.3$

$m = n \cdot {M}_{r} = 24.3 \left(102506.2035\right) = 2490900.744 k g = 2490.900744 g \approx 2491 k g$