Question

The combustion reaction of ${C_3}{H_8}O$ is represented by this balanced chemical equation: $2{C_3}{H_8}O + 9{O_2} \to 6C{O_2} + 8{H_2}O$. When $6.01 \times {10^{22}}$ molecules of ${O_2}$ and $2.78 \times {10^{23}}$ molecules of ${C_3}{H_8}O$ are combined, how many grams of which reagent is used?

Answer 1

Hint: Some chemical reactions contain more than one reactant which are present in different quantities. As the reaction proceeds consumption of any reagent will cause a stoppage of reaction. The reagent which is exhausted first is known as the limiting reagent.

Complete answer:
Organic molecules undergo the process of combustion to form carbon dioxide and water molecules. ${C_3}{H_8}O$ chemically is propanol which undergoes combustion with nine moles of oxygen molecule ${O_2}$to form six mole of $C{O_2}$ and eight moles of ${H_2}O$.
From the balanced chemical reaction we see that for every one mole of ${C_3}{H_8}O$ nine moles of ${O_2}$ is required. Hence, the ratio of these two reagents is $2:9$.
It means that if $2.78 \times {10^{23}}$ molecules of ${C_3}{H_8}O$ is used then according to ratio $8.28 \times {10^{25}}$ molecules of ${O_2}$ must be required.
As from the question we see that we have a smaller number of oxygen molecules than what we require for complete reaction with ${C_3}{H_8}O$. Hence, oxygen is exhausted first and acts as a limiting reagent for the chemical reaction. So, we have to calculate the value of all reagents with respect to oxygen.
We have $6.01 \times {10^{22}}$ molecules of ${O_2}$ which when undergoing reaction require ${C_3}{H_8}O$ in a ratio of $2:9$.
Hence when all the $6.01 \times {10^{22}}$ are used, $1.336 \times {10^{22}}$ molecules of ${C_3}{H_8}O$ are required for complete reaction.
Hence, we see that some molecules of ${C_3}{H_8}O$ remain unreacted during the chemical reaction-
Total molecules of ${C_3}{H_8}O$ remaining after the reaction $ = $ total molecules of ${C_3}{H_8}O$$ - $ molecules used during the chemical reaction
Total molecules of ${C_3}{H_8}O$ remaining after the reaction $ = $$2.78 \times {10^{23}}$$ - $$1.336 \times {10^{22}}$
Total molecules of ${C_3}{H_8}O$ remaining after the reaction$ = $$2.642 \times {10^{23}}$
Mass of particle equal to Avogadro number is equal to atomic mass or molecular mass of the compound
For oxygen:
$6.022 \times {10^{24}}$ molecules of ${O_2}$ $ = $ $32g$${O_2}$
Hence one weight of one molecule of ${O_2}$ is
$1$ molecules of ${O_2}$ $ = $ $\dfrac{{32}}{{6.022 \times {{10}^{24}}}}g$${O_2}$

$6.01 \times {10^{22}}$ molecules of ${O_2}$ $ = $ $\dfrac{{6.01 \times {{10}^{22}}}}{{6.022 \times {{10}^{24}}}} \times 32g$${O_2}$
After the calculation, we find that
$6.01 \times {10^{22}}$ molecules of ${O_2}$$ = $ $0.319g$${O_2}$
Hence, $0.319g$${O_2}$ is required in the combustion process.
For ${C_3}{H_8}O$:
$6.022 \times {10^{24}}$ molecules of ${C_3}{H_8}O$$ = $ $60g$${C_3}{H_8}O$
Hence one weight of one molecule of ${C_3}{H_8}O$ is
$1$ molecules of ${C_3}{H_8}O$ $ = $ $\dfrac{{60}}{{6.022 \times {{10}^{24}}}}g$${C_3}{H_8}O$
We already discussed that $1.336 \times {10^{22}}$ molecules of ${C_3}{H_8}O$ are utilized during the reaction:
$1.336 \times {10^{22}}$ molecules of ${C_3}{H_8}O$ $ = $ $\dfrac{{1.336 \times {{10}^{22}}}}{{6.022 \times {{10}^{24}}}} \times 60g$${C_3}{H_8}O$
After the calculation, we find that
$1.336 \times {10^{22}}$ molecules of ${C_3}{H_8}O$$ = $ $0.132g$${C_3}{H_8}O$
$ \Rightarrow $ After calculation we find that for the combustion reaction $2{C_3}{H_8}O + 9{O_2} \to 6C{O_2} + 8{H_2}O$, $0.319g$${O_2}$ and $0.132g$${C_3}{H_8}O$ are required.

Note:
In chemistry stoichiometry is used to relate quantities of different reactants and products in a chemical reaction. In a chemical reaction whole numerical digits are used to write the balanced chemical equation. Stoichiometry helps in identifying the concentration of the component from the chemical equation.