Answer:
50
Explanation:
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.
Mᵣ: 30.01 32.00 46.01
2NO + O₂ ⟶ 2NO₂
Mass/g: 80.00 16.00
2. Calculate the moles of each reactant
[tex]\text{moles of NO} = \text{80.00 g NO} \times \dfrac{\text{1 mol NO}}{\text{30.01 g NO}} = \text{2.666 mol NO}\\\\\text{moles of O}_{2} = \text{16.00 g O}_{2} \times \dfrac{\text{1 mol O}_{2}}{\text{32.00 g O}_{2}} = \text{0.5000 mol O}_{2}[/tex]
3. Calculate the moles of NO₂ we can obtain from each reactant
From NO:
The molar ratio is 2 mol NO₂:2 mol NO
[tex]\text{Moles of NO}_{2} = \text{2.333 mol NO} \times \dfrac{\text{2 mol NO}_{2}}{\text{2 mol NO}} = \text{2.333 mol NO}_{2}[/tex]
From O₂:
The molar ratio is 2 mol NO₂:1 mol O₂
[tex]\text{Moles of NO}_{2} = \text{0.5000 mol O}_{2}\times \dfrac{\text{2 mol NO}_{2}}{\text{1 mol Cl}_{2}} = \text{1.000 mol NO}_{2}[/tex]
4. Identify the limiting and excess reactants
The limiting reactant is O₂ because it gives the smaller amount of NO₂.
The excess reactant is NO.
5. Mass of excess reactant
(a) Moles of NO reacted
The molar ratio is 2 mol NO:1 mol O₂
[tex]\text{Moles reacted} = \text{0.500 mol O}_{2} \times \dfrac{\text{2 mol NO}}{\text{1 mol O}_{2}} = \text{1.000 mol NO}[/tex]
(b) Mass of NO reacted
[tex]\text{Mass reacted} = \text{1.000 mol NO} \times \dfrac{\text{30.01 g NO}}{\text{1 mol NO}} = \text{30.01 g NO}[/tex]
(c) Mass of NO remaining
Mass remaining = original mass – mass reacted = (80.00 - 30.01) g = 50 g NO
