Question #70307

2 Answers
Stefan V.
Dec 17, 2014

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#!

Michael
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_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#

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