Answer:
[tex]\boxed{\text{520 g H$_{2}$O}}[/tex]
Explanation:
(a) Unbalanced equation
[tex]\rm LiOH + CO$_{2} \longrightarrow \,$ Li$_{2}$CO$_{3}$ + H$_{2}$O[/tex]
(b) Balanced equation
[tex]\rm 2LiOH + CO$_{2} \longrightarrow \,$ Li$_{2}$CO$_{3}$ + H$_{2}$O[/tex]
(c) Molar masses
The only compound for which you need an atomic mass is water.
You look up the atomic masses of each element in the Periodic Table, multiply by their subscripts in the formula, and add.
[tex]\begin{array}{rcrcr}\text{2H} & = & 2 \times 1.008 & = & 2.02\\\text{1O} & = & 1 \times 16.00 & = & 16.00\\\text{H$_{2}$O} & = & & & \mathbf{18.02}\\\end{array}[/tex]
(d) Molar Ratio
You want to convert moles of carbon dioxide to moles of water.
The balanced equation tells you that the molar ratio is 1 mol H₂O:1 mol CO₂.
(e) Significant figures
The only measurement you are given is "about 650 L." That tells you that the trailing zero is not significant.
The volume has only two significant figures, so the mass of water can have only two significant figures.
Note: Intermediate calculations should carry at least one extra digit (a guard digit) to prevent cumulative round-off errors, answers to be reported must have the correct number of significant figures.
(f) The calculation
You don't give the temperature and pressure of the gas, so I shall assume STP (1 bar and 0 °C). At STP, the molar volume of a gas
is 22.71 L.
[tex]\text{Moles of CO$_{2}$} = \text{650 L CO$_{2}$} \times \dfrac{\text{1 mol CO$_{2}$}}{\text{22.71 L CO$_{2}$ }} = \text{28.6 mol CO$_{2}$}\\\\\text{Moles of H$_{2}$O} = \text{28.6 mol CO$_{2}$} \times \dfrac{\text{1 mol H$_{2}$O}}{\text{1 mol CO$_{2}$}} = \text{28.6 mol H$_{2}$O}\\\\\text{Mass of H$_{2}$O} = \text{28.6 mol H$_{2}$O} \times \dfrac{\text{18.02 g H$_{2}$O}}{\text{1 mol H$_{2}$O}} = \boxed{\textbf{520 g H$_{2}$O}}[/tex]
