KCET 2018 Chemistry Solved Paper

For the reaction
2Mg+O2→2MgO
48 g of Mg requires 32 g of O2
Therefore, 1 g of Mg require

32

48

g of O2=0.66g
But available O2 is 0.28g
So, O2 is the limiting reagent and Mg is the excess reagent.
32g of O2 needs 48g of Mg
1g of O2 needs

48

32

g of Mg
0.28 of O2 needs

48

32

×0.28=0.42g
Mg available = 1 g
Mg left =1−0.42=0.58g