Step-by-step explanation:
CaCO₃ ----> CO₂ (g) + CaO
We have to find the grams of calcium carbonate needed to from 3.45 L of CO₂.
First we can find the number of moles of CO₂ that we have to produce. One mol of any gas at STP occupies 22.4 L. We can use this relationship to find the number of moles present in 3.45 L of CO₂.
1 mol of CO₂ = 22.4 L
moles of CO₂ = 3.45 L * 1 mol/(22.4 L)
moles of CO₂ = 0.154 mol
Once we found the number of moles of CO₂, we can find the number of moles of CaCO₃ that are necessary to produce them. According to the coefficients of the reaction, 1 mol of CO₂ will be produced by 1 mol of CaCO₃.
1 mol of CaCO₃ = 1 mol of CO₂
moles of CaCO₃ = 0.154 moles of CO₂ * 1 mol of CaCO₃/(1 mol of CO₂)
moles of CaCO₃ = 0.154 moles
Finally we can convert these moles into grams using the molar mass of CaCO₃.
molar mass of Ca = 40.08 g/mol
molar mass of C = 12.01 g/mol
molar mass of O = 16.00 g/mol
molar mass of CaCO₃ = 1 * 40.08 g/mol + 1 * 12.01 g/mol + 3 * 16.00 g/mol
molar mass of CaCO₃ = 100.09 g/mol
mass of CaCO₃ = 0.154 moles * 100.09 g/mol
mass of CaCO₃ = 15.4 g
Answer: 15.4 g of CaCO₃ will be needed.