can someone help please

Chemistry
Tutor: None Selected Time limit: 1 Day

In an experiment 6.910 g of magnesium is reacted with 6.734 g of oxygen gas according to the equation: 

Mg + O2 ==> MgO. 

A. What is the limiting reactant? 
B. What mass of the other reactant is in exess? 
C. What mass of MgO is produced?
Dec 11th, 2014

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.

Dec 11th, 2014

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