Question

Be sure to answer all parts. Nitric oxide (NO) reacts with oxygen gas to form nitrogen dioxide (NO2), a dark brown gas: 2NO(g) + O2(g) → 2NO2(g)
In one experiment, 0.857 mol of NO is mixed with 0.498 mol of O2.
Determine which of the two reactants is the limiting reactant. Calculate also the number of moles of NO2 produced. Limiting reactant: Moles of NO2 produced: moles

Answer 1

Answer: NO is the limiting reagent in the given reaction and 0.857 moles of $NO_2$ will be produced.

Explanation:

Limiting reagent is defined as the reagent which is present in less amount and it limits the formation of products.

Excess reagent is defined as the reagent which is present in large amount.

For the given chemical reaction:

$2NO(g)+O_2(g)\rightarrow 2NO_2(g)$

We are given:

Moles of NO = 0.857 mol

Moles of oxygen = 0.498 mol

By stoichiometry of the reaction:

If 2 moles of NO reacts with 1 mole of oxygen gas.

So, 0.857 moles of NO will react with = $\frac{1}{2}\times 0.857=0.4285mol$ of $O_2$

As, the given amount of oxygen gas is more than the required amount. So, it is considered as an excess reagent.

Thus, NO is considered as the limiting reagent because it limits the formation of products.

By Stoichiometry of the reaction:

If 2 moles of NO produces 2 moles of nitrogen dioxide gas.

So, 0.857 moles of NO will produce = $\frac{2}{2}\times 0.857=0.857mol$ of $NO_2$

Hence, NO is the limiting reagent in the given reaction and 0.857 moles of $NO_2$ will be produced.