# Question 067bf

Apr 14, 2016

Here's what I got.

#### Explanation:

Your starting point here will be the balanced chemical equation for the combustion of acetylene, ${\text{C"_2"H}}_{2}$, which looks like this

$\textcolor{g r e e n}{2} {\text{C"_ 2"H"_ (2(g)) + color(red)(5)"O"_ (2(g)) -> color(blue)(4)"CO"_ (2(g)) + color(purple)(2)"H"_ 2"O}}_{\left(l\right)}$

Notice that the two reactants take part in the reaction in a $\textcolor{g r e e n}{2} : \textcolor{red}{5}$ mole ratio. Keep this in mind.

Now, use the molar masses of acetylene and of oxygen gas to determine how many moles of each you're mixing

5.00 color(red)(cancel(color(black)("g"))) * ("1 mole C"_2"H"_2)/(26.04color(red)(cancel(color(black)("g")))) = "0.1920 moles C"_2"H"_2

1.00 color(red)(cancel(color(black)("g"))) * "1 mole O"_2/(32.0color(red)(cancel(color(black)("g")))) = "0.03125 moles O"_2

At this point, it's clear that oxygen gas will act as a limiting reagent. The reaction requires that you have $\frac{\textcolor{red}{5}}{\textcolor{g r e e n}{2}}$ times more moles of oxygen gas than of acetylene, but the mixture actually contains less oxygen gas than acetylene $\to$ oxygen as will be the limiting reagent.

So, oxygen gas will be completely consumed by the reaction before all the moles of acetylene get a chance to react. More specifically, only

0.03125 color(red)(cancel(color(black)("moles O"_2))) * (color(green)(2)color(white)(a)"moles C"_2"H"_2)/(color(red)(5)color(red)(cancel(color(black)("moles O"_2)))) = "0.0125 moles C"_2"H"_2

will actually take part in the reaction, the rest will be in excess. Once the reaction is complete, the reaction vessel will contain

${n}_{{C}_{2} {H}_{2}} = \text{0.1920 moles" - "0.0125 moles}$

$= {\text{0.1795 moles C"_2"H}}_{2}$

This will be equivalent to

$0.1795 \textcolor{red}{\cancel{\textcolor{b l a c k}{{\text{moles C"_2"H"_2))) * "26.04 g"/(1color(red)(cancel(color(black)("mole C"_2"H"_2)))) = color(green)(|bar(ul(color(white)(a/a)"4.67 g C"_2"H}}_{2} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

The oxygen gas will be completely consumed by the reaction, so you'll be left with

${m}_{{O}_{2}} = \textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} {\text{0 g O}}_{2} \textcolor{w h i t e}{\frac{a}{a}} |}}} \to$ completely consumed by the reaction

Now, to make calculations easier, pick the number of moles of acetylene that take part in the reaction and use the $\textcolor{g r e e n}{2} : \textcolor{b l u e}{4}$ mole ratio that exists between acetylene and carbon dioxide to find how many moles of the latter are produced by the reaction

0.0125color(red)(cancel(color(black)("moles C"_2"H"_2))) * (color(blue)(4)color(white)(a)"moles CO"_2)/(color(green)(2)color(red)(cancel(color(black)("moles C"_2"H"_2)))) = "0.0250 moles CO"_2

Use carbon dioxide's molar mass to convert this to grams

$0.0250 \textcolor{red}{\cancel{\textcolor{b l a c k}{{\text{moles CO"_2))) * "44.01 g"/(1color(red)(cancel(color(black)("mole CO"_2)))) = color(green)(|bar(ul(color(white)(a/a)"1.10 g CO}}_{2} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

Do the same for water.

0.0125 color(red)(cancel(color(black)("moles C"_2"H"_2))) * (color(purple)(2)color(white)(a)"H"_2"O")/(color(green)(2)color(red)(cancel(color(black)("moles C"_2"H"_2)))) = "0.0125 moles H"_2"O"#

This will be equivalent to

$0.0125 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{moles H"_2"O"))) * "18.015 g"/(1color(red)(cancel(color(black)("mole H"_2"O")))) = color(green)(|bar(ul(color(white)(a/a)"0.225 g H"_2"O} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

The values are rounded to three sig figs.

Now, double-check the results by using the principle of mass conservation. The initial mass of the acetylene + oxygen gas mixture was

${m}_{\text{initial" = "5.00 g" + "1.00 g" = "6.00 g}}$

If you exclude the mass of unreacted acetylene, you will be left with the mass of reactants that actually took part in the reaction

${m}_{\text{reactants" = "6.00 g" - "4.67 g" = "1.33 g}}$

Now, the total mass of the reactants must be equal to the total mass of the products, which is

${m}_{\text{products" = "1.10 g" + "10.225 g" = "1.325 g" ~~ "1.33 g}}$

The values are not an exact math because of rounding.

So, the total mass of chemical species present in the sample is unchanged by the reaction. You start with $\text{6.00 g}$ of acetylene and oxygen gas and end up with

$\text{4.67 g " + " 1.325 g" = "5.995 g" ~~ "6.00 g}$

of acetylene, carbon dioxide, and water. The oxygen gas is no longer present after the reaction is complete.