# NASA used canisters filled with lithium hydroxide to remove carbon dioxide inside the space shuttles. The chemical equation for the reaction is: "2LiOH(s)+CO"_2("g")"rarr"Li"_2"CO"_3("s")+"H"_2"O(g)" ?

## a. Determine the theoretical mass of lithium hydroxide required for each 5.00 g of carbon dioxide removed from the shuttle b. Theoretically, what mass of lithium carbonate, ${\text{Li"_2"CO}}_{3}$, would be produced for each 5.00 g of carbon dioxide removed from the shuttle?

Nov 6, 2017

Here's what I got.

#### Explanation:

The first thing that you need to do here is to convert the mass of carbon dioxide to moles by using the molar mass of the compound.

This will allow you to use the $2 : 1$ mole ratio that exists between the two reactants in the balanced chemical equation to find the number of moles of lithium hydroxide needed to remove that much carbon dioxide from the shuttle.

5.00 color(red)(cancel(color(black)("g"))) * "1 mole CO"_2/(44.01color(red)(cancel(color(black)("g")))) = "0.1136 moles CO"_2

In order for the reaction to consume $0.1136$ moles of carbon dioxide, it must also consume

0.1136 color(red)(cancel(color(black)("moles CO"_2))) * overbrace("2 moles LiOH"/(1color(red)(cancel(color(black)("mole CO"_2)))))^(color(blue)("given by the balanced chemical equation")) = "0.2272 moles LiOH"

To convert this to grams of lithium hydroxide, use the molar mass of the compound.

$0.2272 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{moles LiOH"))) * "23.95 g"/(1color(red)(cancel(color(black)("mole LiOH")))) = color(darkgreen)(ul(color(black)("5.44 g}}}}$

The answer is rounded to three sig figs.

In order to find the mass of lithium carbonate that can be produced for every $\text{5.00 g}$ of carbon dioxide removed from the shuttle, use the $1 : 1$ mole ratio that exists between the two compounds to say that theoretically, when the reaction consumes $0.1136$ moles of carbon dioxide, it produces $0.1136$ moles of lithium carbonate.

Once again, to convert the number of moles to grams, use the molar mass of the compound.

$0.1136 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{moles Li"_2"CO"_3))) * "73.891 g"/(1 color(red)(cancel(color(black)("mole Li"_2"CO"_3)))) = color(darkgreen)(ul(color(black)("8.39 g}}}}$

The answer is rounded to three sig figs.