There are 70 m^3 of calcium hydroxide and a gas mixture that is 7% CO2. What volume of CO2 is needed to convert the calcium hydroxide to CaCO3?

Calcium hydroxide (Ca(OH)2) is converted to calcium carbonate (CaCO3) by treating it with carbon dioxide (CO2).  The equation is shown below:

Ca(OH)2 + CO2 --> CaCO3 + H2O

We have 70 m3 (cubic meters) of calcium hydroxide.  If we convert this to milliliters (also called cubic centimeters), then we can use the density of calcium hydroxide (2.211 g/mL) to find the mass in grams.

70 m3 * (1 x 10e6 mL / m3) * (2.211 g / mL) = 1.55 x 10e8 g Ca(OH)2

Now divide by the molecular weight of Ca(OH)2 to get the amount in moles.

1.55 x 10e8 g * (1 mole / 74.093 g) = 2,091,965.5 moles Ca(OH)2

We can see from the chemical equation above that one mole of CO2 will react with one mole of Ca(OH)2, so we will need the exact same number of moles of CO2.  The molar volume of any ideal gas is 22.414 L per mole, so we can use this as a conversion factor to convert the moles of CO2 into a volume in liters.

2,091,965.5 moles * (22.414 L / mole) = 46,889,314.7 L CO2

So that is the amount of pure CO2 gas needed for the reaction.  Since you stated that the gas on hand is 7% CO2, divide the volume by 0.07 to find the volume of that particular gas mixture required for the reaction.

46,889,314,7 L / 0.07 = 669,847,353.1 liters of gas