help please! 1.50 g of NH3 reacts with 2.75 g of O2.?

Question

a) which is the limiting reactant?
b) how many grams of NO and H2O form?
c) how many grams of the excess reactant remain after the limiting reactant is consumed?

Answer 1

The balanced equation is
4 NH3 + 5 O2 >> 4 NO + 6H2O

MM NH3 = 17 g/mol
1.50 g / 17 = 0.0882 mole NH3

MM O2 = 32 g/mol
2.75 / 32 = 0.0859 mole O2
the ratio between NH3 and O2 is 4 : 5 so O2 is the limiting reactant

The ratio between O2 ( limiting reactant ) and NO is 5 : 4
5 : 4 = 0.0859 : x
x = mole NO = 0.0688
MM = 30 g/mol
0.0688 mol x 30 g/mol = 2.06 g NO formed

the ratio between O2 and H2O is 5 : 6
5 : 6 = 0.0859 : x
x = 0.103 mole H2O
MM = 18 g/mol
0.103 x 18 = 1.86 g H2O formed

the ratio between NH3 and O2 is 4 : 5
4 : 5 = x : 0.0859
x = 0.0687 mole NH3 needed for the reaction
0.0882 - 0.0687 = 0.0195 mole NH3 in excess
0.0195 x 17 = 0.332 g of NH3 in excess

Answer 2

4 NH3 + 5 O2 -> 4 NO + 6 H2O

GMM of NH3 = 17

GMM of H2O = 18