# Question #70307

Dec 17, 2014

I'm a little confused by this...

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

Let's assume that an initial mass of $x$ grams of $M g$ reacted with the ${O}_{2}$ according to the balanced chemical reaction. If the ${O}_{2}$ was neither in excess, nor a limiting reagent (reacted completely), then any mass of $M {g}_{\left(s\right)}$ added would automatically be in excess IF:

1. Initially, ${O}_{2}$ was not in excess;
2. The mass of ${O}_{2}$ was not supplimented when the mass of $M g$ was.

If ${O}_{2}$ was a limiting reagent before the added mass of $M g$, then the mass of $M g$ that will react will be equal to the initial mass of $x$ grams.

So, my best answer is DEPENDS ON THE ${O}_{2}$!

Dec 17, 2014

0.2g of Mg has reacted.

If the mass of Mg has increased by 0.335g then this must be equal to the mass of MgO.

${M}_{r} M g O = \left(24 + 16\right) = 40$

So 1 mole weighs 40g

So no. moles MgO = $\frac{m}{M} _ r = \frac{0.335}{40} = 0.00835$

So no. moles Mg = 0.00835 since 1 mole MgO contains 1 mole Mg.

So mass Mg$= n \times {A}_{r} = 0.00835 \times 24 = 0.2 g$