N is 14 g/mol
H is 1 g/mol
O is 16 g/mol
so
NH3 is 14 + 3 = 17 g/mol
O2 is 32 g/mpl
you need 7 mols of O2 for every 4 mols of NH3
10 g NH3 = 10 g ( 1 mol / 17 g) = 0.588 mols
so we need
0.588 mol *7/4 = 1.03 mols of O2
1.03 mols* 32 g/mol = 32.9 grams of O2
WHOOOPS we only have 25.3 grams of O2
so
O2 is limiting reagent
25.3 grams O2 * 1 mol/ 32 g = 0.79 mols O2 ----> Use that
for every 7 mols O2 we 6 mols H2O
so mols H2O = (6/7)* .79 = 0.677 mols H2O produced
H2O = 2 + 16 = 18 g/mol
0.677 *18 = 12.2 grams H2O
4nh3 +7o2 -->4no2+6h20. what mass of h20 forms when 10.0 g of nh3 reacts with 25.3g of o2
1 answer