If 16.00 g of O₂ reacts with 80.00 g NO, how many the excess reactant are left over? (enter only the value, round to whole number)

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  

3. Calculate the moles of NO₂ we can obtain from each reactant

From NO:

The molar ratio is 2 mol NO₂:2 mol NO

From O₂:

The molar ratio is 2 mol NO₂:1 mol O₂

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

(b) Mass of NO reacted

(c) Mass of NO remaining

Mass remaining = original mass – mass reacted = (80.00 - 30.01) g = 50 g NO

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