Question

Question #70307

Answer 1

I'm a little confused by this...

`2Mg_((s)) + O_2(g) -> 2MgO_((s))`

Let's assume that an initial mass of `x` grams of `Mg` 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 `Mg_((s))` 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 `Mg` was.

If `O_2` was a limiting reagent before the added mass of `Mg`, then the mass of `Mg` that will react will be equal to the initial mass of `x` grams.

So, my best answer is DEPENDS ON THE `O_2`!

Answer 2

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_rMgO=(24+16)=40`

So 1 mole weighs 40g

So no. moles MgO = `m/M_r=0.335/40=0.00835`

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

So mass Mg`= nxxA_r=0.00835xx24=0.2g`