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If 3.5 moles of nitrogen monoxide (NO) react with 6.0 moles of oxygen gas (O2), how many moles of the product can be formed and how many moles of the excess reactant will be left over when the reaction is complete? Show all of your work. unbalanced equation: NO + O2 “yields”/ NO2

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isyllus

Answer: -

4.25 mol of Oâ‚‚ left as excess.

3.5 mol of NOâ‚‚ formed.

Explanation: -

Number of moles of NO taken = 3.5

Number of moles of Oâ‚‚ taken = 6.0

The balanced chemical equation for this reaction is

2 NO+ O₂ → 2 NO₂

From the equation we can see that

2 mol of NO react with 1 mol of Oâ‚‚

3.5 mol of NO react with [tex] \frac{1 mol O2}{2 mol NO} [/tex] x 3.5 mol NO

= 1.75 mol Oâ‚‚

So Oxygen Oâ‚‚ is in excess and NO is the limiting reagent.

Moles of Oâ‚‚ left over = 6 - 1.75 =4.25 mol of Oâ‚‚

From the balanced chemical equation we see

2 mol of NO gives 2 mol of NOâ‚‚

3.5 mol of NO gives[tex] \frac2 mol NO2}{2 mol NO} [/tex] x 3.5 mol NO

= 3.5 mol of NOâ‚‚

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