# Question 39fd9

Apr 23, 2016

Oxygen gas will be the limiting reagent here.

#### Explanation:

Start by taking a look at the balanced chemical equation that describes the combustion of carbon disulfide, ${\text{CS}}_{2}$

${\text{CS"_ (2(l)) + color(red)(3)"O"_ (2(g)) -> "CO"_ (2(g)) + color(blue)(2)"SO}}_{2 \left(g\right)}$

For starters, carbon disulfide and oxygen gas react in a $1 : \textcolor{red}{3}$ mole ratio, which means that the reaction will always consume three times as many moles of oxygen gas than moles of carbon disulfide.

You are told that you have $1.35$ moles of carbon disulfide and $0.63$ moles of oxygen gas, which makes it clear, since you actually have fewer moles of oxygen gas, that oxygen gas will act as a limiting reagent.

This means that the oxygen gas will be completely consumed by the reaction before all the moles of carbon disulfide will get the chance to react $\to$ carbon disulfide will be in excess.

More specifically, the reaction will consume only

0.63 color(red)(cancel(color(black)("moles O"_2))) * "1 mole CS"_2/(color(red)(3)color(red)(cancel(color(black)("moles O"_2)))) = "0.21 moles CS"_2

The rest of the available carbon disulfide will be in excess, i.e. remain unreacted.

n_(CS_2) = "1.35 moles" - "0.21 moles" = color(green)(|bar(ul(color(white)(a/a)"1.14 moles CS"_2color(white)(a/a)|)))

To find the mass of carbon disulfide that is in excess, use the compound's molar mass

1.14 color(red)(cancel(color(black)("moles CS"_2))) * "76.141 g"/(1color(red)(cancel(color(black)("mole CS"_2)))) = "86.8 g"

So, you know that your reaction will consume

• ${\text{0.63 moles O}}_{2} \to$ completely consumed
• ${\text{0.21 moles CS}}_{2}$

Pick one of the two reactants and use the mole ratios that exist between itself and the two products to find the number of moles of carbon dioxide and sulfur dioxide produced by the reaction.

To make the calculations easier, I'll pick carbon disulfide. You will have

$0.21 \textcolor{red}{\cancel{\textcolor{b l a c k}{{\text{moles CS"_2))) * "1 mole CO"_2/(1color(red)(cancel(color(black)("mole CS"_2)))) = color(green)(|bar(ul(color(white)(a/a)"0.21 moles CO}}_{2} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

In grams, this would be

0.21 color(red)(cancel(color(black)("moles CO"_2))) * "44.01 g"/(1color(red)(cancel(color(black)("mole CO"_2)))) = "9.2 g"

For sulfur dioxide, you'll have

$0.21 \textcolor{red}{\cancel{\textcolor{b l a c k}{{\text{moles CS"_2))) * (color(blue)(2)color(white)(a)"moles SO"_2)/(1color(red)(cancel(color(black)("mole CS"_2)))) = color(green)(|bar(ul(color(white)(a/a)"0.42 moles SO}}_{2} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

In grams, this would be

0.42 color(red)(cancel(color(black)("moles SO"_2))) * "64.06 g"/(1color(red)(cancel(color(black)("mole SO"_2)))) = "27 g"#