1. O2 is the limiting reagent
2. Excess NH3 = 0.102 moles
3. 0.088 mol of NO is formed
4. 2.376 grams of H2O is formed
Stoichiometric problem
From the equation: [tex]4NH_3 + 5O_2 --- > 4NO + 6H_2O[/tex]
Mole ratio of NH3 and O2 = 4:5
Mole of 3.25 g NH3 = 3.25/17 = 0.19 mol
Mole of 3.50 g O2 = 3.5/32 = 0.11 mol
Equivalent mole of O2 from NH3 = 5/4 x 0.19 = 0.24 mol
Equivalent mole of NH3 from O2 = 4/5 x 0.11 = 0.088 mol
Thus, O2 is the limiting reagent.
Excess NH3 = 0.19 - 0.088 = 0.102 moles
Mole ratio of O2 and NO = 5:4
Equivalent mole of NO = 4/5 x 0.11 = 0.088 mol
Mole ratio of O2 and H2O = 5:6
Equivalent mole of H2O = 6/5 x 0.11 = 0.132 mol
Mass of 0.132 mol H2O = 0.132 x 18 = 2.376 grams
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