In an experiment 6.910 g of magnesium is reacted with 6.734 g of oxygen gas according to the equation:
Divide by molar mass to find # of moles of each. O2 is 32, Mg is 12, MgO is 28.
Balance the equation: 2Mg + O2 = 2MgO
Just offhand, the limiting reactant is O2 since there are less moles of it, and 6.734/32 moles of O2 are consumed in this reaction.
Let me go get a calculator...
6.734/32 = 0.2104 moles of O2 reacted to produce twice as many moles of MgO. So multiply by 28 and then by 2 to get 11.78 g of MgO produced.
0.4208 moles of Mg is 5.051g, which reacted; the rest did not react. Thus, the leftover Mg is 6.910-5.051 = 1.859g.
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