Calculate the mass of CaCO3 that would produce 40 mL of CO2 at STP....Please show calculation...
3 Answers

First, we need to write a chemical equation for this reaction. When heated, calcium carbonate decomposes to produce calcium oxide and carbon dioxide according to the balanced equation:
CaCO3(s) → CaO(s) + CO2(g)
Next, we use dimensional analysis to convert 40. mL of CO2 at STP to liters of CO2 at STP, to moles of CO2 at STP, to moles of CaCO3, to grams of CaCO3. To do this, we need the following equalities to construct conversion factors:
1 L CO2 = 1,000 mL CO2
1 mol CO2 = 22.4 L CO2
1 mol CO2 = 1 mol CaCO3
1 mol CaCO3 = 100.088 g CaCO3
[(49. mL CO2)/1][(1 L CO2)/(1,000 mL CO2)][(1 mol CO2)/(22.4 L CO2)][(1 mol CaCO3)/(1 mol CO2)][(100.088 g CaCO3)/(1 mol CaCO3)] = 0.1787271 g CaCO3 = 0.18 g CaCO3 to two significant figures
Answer: The mass of CaCO3 that would produce 40. mL of CO2 gas at STP is 0.18 grams.
Source(s): Periodic Table of the Elements
personal knowledge 
CaCO3 + 2HCl > CaCl2 + H2O + CO2
No of moles of CO2
= Volume / Molar Volume
= 40 / 24000
= (1 / 600) mol
From the equation
1 mol of CO2 is produced from 1 mol of CaCO3
Therefore, (1 / 600) mol of CO2 is produced from (1 / 600) mol of CaCO3
Mass of CaCO3
= Number of moles x Molar mass
= (1 / 600) x 100.09
= 0.17 g

500gCaO * 1 molCaO /56g CaO * 1mol CaCO3/1molCaO(mole ratio from the reaction) * 100g CaCO3/1molCaCO3 =893gCaCO3