The reaction forms 2.63 g NO.
We have the masses of two reactants, so this is a limiting reactant problem.
We know that we will need a balanced equation with masses, moles, and molar masses of the compounds involved.
Step 1. Gather all the information in one place with molar masses above the formulas and everything else below them.
M_r: ____17.03 __32.00 _30.01
_______4NH_3 + 5O_2 → 4NO + 6H_2O
Mass/g: _3.25 ___3.50
Step 2. Calculate the moles of each reactant
Moles of NH_3 = 3.25 g NH_3 × (1 mol NH_3 /17.03 g NH_3)
= 0.1908 mol NH_3
Moles of H_2 = 3.50 g O_2 × (1 mol O_2/32.00 g O_2) = 0.1094 mol O_2
Step 3. Identify the limiting reactant
Calculate the moles of NO we can obtain from each reactant.
From NH_3: Moles of NO = 0.1908 mol NH_3 × (4 mol NO/4 mol NH_3)
= 0.1908 mol NO
From O_2: Moles of NO = 0.1094 mol O_2 × (4 mol NO /5 mol O_2)
= 0.087 50 mol NO
O_2 is the limiting reactant because it gives the smaller amount of NO.
Step 4. Calculate the mass of NO
Mass of NO = 0.087 50 mol NO × (30.01g NO/1 mol NO) = 2.63 g NO
