Answer:
a) 1.39 g ; b) O₂ is limiting reactant,  NH₃ is excess reactant; c) 0.7 g
Explanation:
We have the masses of two reactants, so this is a limiting reactant problem.
We will need a balanced equation with masses, moles, and molar masses of the compounds involved.
1. Gather all the information in one place with molar masses above the formulas and masses below them. Â
MM: Â Â Â Â 17.03 Â Â 32.00 Â Â 30.01
       4NH₃  +  5O₂ ⟶ 4NO + 6H₂O
Mass/g: Â Â 1.5 Â Â Â Â 1.85
2. Calculate the moles of each reactant Â
[tex]\text{moles of NH}_{3} = \text{1.5 g NH}_{3} \times \dfrac{\text{1 mol NH}_{3}}{\text{17.03 g NH}_{3}} = \text{0.0881 mol NH}_{3}\\\\\text{moles of O}_{2} = \text{1.85 g O}_{2} \times \dfrac{\text{1 mol O}_{2}}{\text{32.00 g O}_{2}} = \text{0.057 81 mol O}_{2}[/tex]
3. Calculate the moles of NO we can obtain from each reactant
From NH₃:
The molar ratio is 4 mol NO:4 mol NH₃
[tex]\text{Moles of NO} = \text{0.0881 mol NH}_{3} \times \dfrac{\text{4 mol NO}}{\text{4 mol NH}_{3}} = \text{0.0881 mol NO}[/tex]
From Oâ‚‚:
The molar ratio is 4 mol NO:5 mol Oâ‚‚
[tex]\text{Moles of NO} = Â \text{0.057 81 mol O}_{2}\times \dfrac{\text{4 mol NO}}{\text{5 mol O}_{2}} = \text{0.046 25 mol NO}[/tex]
4. Identify the limiting and excess reactants
The limiting reactant is Oâ‚‚ because it gives the smaller amount of NO.
The excess reactant is NH₃.
5. Calculate the mass of NO formed
[tex]\text{Mass of NO} = \text{0.046 25 mol NO}\times \dfrac{\text{30.01 g NO}}{\text{1 mol NO}} = \textbf{1.39 g NO}[/tex]
6. Calculate the moles of NH₃ reacted
The molar ratio is 4 mol NH₃:5 mol O₂
[tex]\text{Moles reacted} = \text{0.057 81 mol O}_{2} \times \dfrac{\text{4 mol NH}_{3}}{\text{5 mol O}_{2}} = \text{0.046 25 mol NH}_{3}[/tex]
7. Calculate the mass of NH₃ reacted
[tex]\text{Mass reacted} = \text{0.046 25 mol NH}_{3} \times \dfrac{\text{17.03 g NH}_{3}}{\text{1 mol NH}_{3}} = \text{0.7876 g NH}_{3}[/tex]
8. Calculate the mass of NH₃ remaining
Mass remaining = original mass – mass reacted = (1.5 - 0.7876) g = 0.7 g NH₃
