If you reacted 88.9 g of ammonia with excess oxygen, what mass of nitric oxide would you expect to make? you will need to balance the equation first. nh3(g) + o2(g) -> no(g) + h2o(g)

1) Balanced chemical equation

4NH3(g) + 5O2(g) ---> 4NO(g) + 6H2O(g)

2) State the molar ratios

4 mol NH3 : 5 mol O2 : 4 mol NO : 6 mol H2O

3) Convert 88.9 g of ammonia to moles, using the molar mass

molar mass of NH3 = 14 g/mol + 3 * 1 g/mol = 17 g/mol

number of moles = mass in grams / molar mass = 88.9 g / 17 g/mol = 5.23 mol NH3

4) Make the proportion

4 mol NO / 4 mol NH3 = x / 5.23 mol NH3=> x = 5.23 mol NO

5) Convert 5.23 mol NO to grams

molar mass NO = 14 g/mol + 16g/mol = 30 g/mol

mass = number of moles * molar mass = 5.23 mol * 30 g/mol = 156.9 g ≈ 157 g

157 grams