The following reaction can be used to remove \(\mathrm{CO}_{2}\) breathed out by astronauts in a spacecraft. $$2 \mathrm{LiOH}(s)+\mathrm{CO}_{2}(g) \rightarrow \mathrm{Li}_{2} \mathrm{CO}_{3}(s)+\mathrm{H}_{2} \mathrm{O}(l)$$ \begin{equation}\begin{array}{l}{\text { a. How many grams of carbon dioxide can }} \\ {\text { be removed by } 5.5 \text { mol LiOH? }} \\ {\text { b. How many milliliters } \mathrm{H}_{2} \mathrm{O} \text { (density }=} \\\ {0.997 \mathrm{g} / \mathrm{mL} \text { ) could form from } 25.7 \mathrm{g} \text { LiOH? }} \\ {\text { c. How many molecules } \mathrm{H}_{2} \mathrm{O} \text { could be }} \\ {\text { made when } 3.28 \mathrm{g} \mathrm{CO}_{2} \text { react? }}\end{array}\end{equation}

Step by step solution

01

Identify and Convert

Firstly, identify the balanced equation given: \(2 \mathrm{LiOH}(s)+\mathrm{CO}_{2}(g) \rightarrow \mathrm{Li}_{2} \mathrm{CO}_{3}(s)+\mathrm{H}_{2} \mathrm{O}(l).\n For part a: Given that 5.5 mol of LiOH reacts, determine the moles of \(\mathrm{CO}_{2}\) using the balanced equation: \(5.5 mol \ LiOH \times \frac{1 mol \ CO_{2}}{2 mol \ LiOH}\).

02

Molar Mass Calculation

Next, calculate the molar mass of \(CO_{2}\) to convert moles of \(CO_{2}\) to grams: \(1 mol \ CO_{2} = 44.01 g \ CO_{2}\). Hence, multiply the result of step 1 with the molar mass of \(CO_{2}\) to find how many grams can be removed.

03

Volume of water formation

For part b: Given that 25.7 g of LiOH react, convert this to moles using its molar mass (23.95 g/mol). Subsequently, using the balanced equation, convert moles of LiOH to moles of \(H_{2}O\). Then convert moles of \(H_{2}O\) to grams using its molar mass (18.02 g/mol). Lastly, convert grams of \(H_{2}O\) to milliliters using the given density.

04

Quantity of water molecules

For part c: Start by converting the given mass of \(CO_{2}\) to moles using its molar mass. Using the balanced equation, convert moles of \(CO_{2}\) to moles of \(H_{2}O\). To determine the number of molecules, use Avogadro's number (\(6.022 \times 10^{23} molecules/mol\)).

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