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Chapter 3: Problem 83

Nitric oxide (NO) reacts with oxygen gas to form nitrogen dioxide \(\left(\mathrm{NO}_{2}\right),\) a dark-brown gas: $$ 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{NO}_{2}(g) $$ In one experiment 0.886 mole of \(\mathrm{NO}\) is mixed with 0.503 mole of \(\mathrm{O}_{2}\). Calculate which of the two reactants is the limiting reagent. Calculate also the number of moles of \(\mathrm{NO}_{2}\) produced.

Short Answer

Expert verified
The limiting reactant is Nitric oxide (NO) and 0.886 moles of Nitrogen dioxide (NO2) will be produced.

Step by step solution

01

Write the Balanced Chemical Equation

The balanced chemical equation of the reaction already given: \(2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{NO}_{2}(g)\)
02

Calculate the stoichiometric ratio

From the balanced equation, we can see that two moles of \(\mathrm{NO}\) reacts with one mole of \(\mathrm{O}_{2}\) to produce two moles of \(\mathrm{NO}_{2}\). Therefore, for every mole of \(\mathrm{O}_{2}\), we need two moles of \(\mathrm{NO}\). And for every two moles of \(\mathrm{NO}\), we get two moles of \(\mathrm{NO}_{2}\).
03

Calculate how much reactant each reactant can consume

We have 0.886 moles of \(\mathrm{NO}\) and 0.503 moles of \(\mathrm{O}_{2}\). If all of the \(\mathrm{NO}\) were to react, it would need 0.443 (0.886/2) moles of \(\mathrm{O}_{2}\). We have more than this amount, so \(\mathrm{NO}\) will get used up first. Thus, \(\mathrm{NO}\) is the limiting reactant.
04

Calculate the moles of product

Now that we know that \(\mathrm{NO}\) is the limiting reactant, we can calculate the amount of \(\mathrm{NO}_{2}\) produced. From the stoichiometric ratio, we know that two moles of \(\mathrm{NO}\) produce two moles of \(\mathrm{NO}_{2}\). Thus, 0.886 moles of \(\mathrm{NO}\) will produce 0.886 moles of \(\mathrm{NO}_{2}\).

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Most popular questions from this chapter

A mixture of \(\mathrm{CuSO}_{4} \cdot 5 \mathrm{H}_{2} \mathrm{O}\) and \(\mathrm{MgSO}_{4} \cdot 7 \mathrm{H}_{2} \mathrm{O}\) is heated until all the water is lost. If \(5.020 \mathrm{~g}\) of the mixture gives \(2.988 \mathrm{~g}\) of the anhydrous salts, what is the percent by mass of \(\mathrm{CuSO}_{4} \cdot 5 \mathrm{H}_{2} \mathrm{O}\) in the mixture?

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