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## Q. 3.21

The results of the Miller–Urey experiment require a chemical reaction to account for the formation of glycine. The reaction between carbon dioxide, ammonia, and methane that produces glycine, water, and carbon monoxide is described by the following chemical equation:

$2 CO_{2}(g) + NH_{3}(g) + CH_{4}(g) \rightarrow C_{2}H_{5}NO_{2}(s) + H_{2}O( \ell ) + CO(g)$

How much glycine could be expected from the reaction of 29.3 g of $CO_{2}$, 4.53 g of $NH_{3}$, and 4.27 g of $CH_{4}$?

## Verified Solution

Collect and Organize We are given a balanced chemical equation and the quantities of three reactants $(CH_{4}, NH_{3}, and CO_{2})$. We are asked how much glycine can be produced, which will require us to identify which one of the reactants is the limiting reactant. Since there are three reactants, we must use the first method discussed previously.

Analyze First we calculate the number of moles of each reactant we have available. Then, we calculate the maximum amount of glycine that would be produced from each reactant. The reactant yielding the smallest amount of glycine is the limiting reactant and determines the maximum amount of glycine that we can expect from the reaction.

Solve

1. Using the molar masses of $CO_{2}, NH_{3}, and CH_{4}$, we calculate the number of moles of each that are available:

$29.3 \sout{g CO_{2}}\times \frac{1 mol CO_{2}}{44.01 \sout{g CO_{2}}}=0.6658 mol CO_{2}$

$4.53 \sout{ g NH_{3}}\times \frac{1 mol NH_{3}}{17.03 \sout{g NH_{3}}}=0.2660 mol NH_{3}$

$4.27 \sout{g CH_{4}}\times \frac{1 mol CH_{4}}{16.04 \sout{g CH_{4}}}=0.2662 mol CH_{4}$

2. Using the mole ratios of $C_{2}H_{5}NO_{2}$ for each reactant and the molar mass of $C_{2}H_{5}NO_{2}$, we calculate the amount of glycine that could be produced from each reactant:

$0.6658 \sout{mol CO_{2}}\times \frac{1 \sout{mol C_{2}H_{5}NO_{2}}}{2 \sout{mol CO_{2}}}\times \frac{75.07 g C_{2}H_{5}NO_{2}}{1 \sout{mol C_{2}H_{5}NO_{2}}}=25.0 g C_{2}H_{5}NO_{2}$

$0.2660 \sout{mol NH_{3}}\times \frac{1 \sout{mol C_{2}H_{5}NO_{2}}}{1 \sout{mol NH_{3}}}\times \frac{75.07 g C_{2}H_{5}NO_{2}}{1 \sout{mol C_{2}H_{5}NO_{2}}}=20.0 g C_{2}H_{5}NO_{2}$

$0.2662 \sout{ mol CH_{4}}\times \frac{1 \sout{mol C_{2}H_{5}NO_{2}}}{1 \sout{mol CH_{4}}}\times \frac{75.07 g C_{2}H_{5}NO_{2}}{1 \sout{mol C_{2}H_{5}NO_{2}}}=20.0 g C_{2}H_{5}NO_{2}$

Of these three possible yields of $C_{2}H_{5}NO_{2}$, we can only make the smallest amount, 20.0 g of $C_{2}H_{5}NO_{2}$, so both ammonia and methane are considered limiting reactants in this reaction.

Think About It Although the mass of methane available for the reaction is the lowest of the three reactants, there is an equal number of moles of methane and ammonia. Since carbon dioxide is present in excess, both methane and ammonia are limiting reactants. With three or more reactants, one or more may turn out to be in excess.