When heated, calcium carbonate decomposes to yield calcium oxide and carbon dioxide gas. How many grams of calcium carbonate are needed to produce 81.0 of carbon dioxide at STP?
Is the asnwer 80.95
The reaction is
CaCO3 -> CaO + CO2
What are the units of your 81.0? Liters? Whatever it is, convert it to moles at STP.
81.0 l is 3.61 moles at STP. You will need the same number of moles of CaCO3.
1 mole of CaCO3 has a mass of 40 + 12 + 48 = 100 g
80.95 g is not the answer.
To find the number of grams of calcium carbonate needed to produce a given amount of carbon dioxide gas, you need to use the balanced chemical equation for the decomposition of calcium carbonate. The balanced equation is:
CaCO3(s) -> CaO(s) + CO2(g)
From the balanced equation, we can see that the mole ratio between CaCO3 and CO2 is 1:1. This means that for each mole of CaCO3, one mole of CO2 is produced.
To calculate the number of moles of CO2, you can use the ideal gas law equation:
PV = nRT
Where:
P = pressure (at STP, the pressure is 1 atm)
V = volume
n = number of moles
R = ideal gas constant (0.0821 L·atm/mol·K)
T = temperature (at STP, the temperature is 273 K)
Since we are given the volume of carbon dioxide gas at STP as 81.0 L, we can solve for the number of moles of CO2:
n = PV / RT = (1 atm)(81.0 L) / (0.0821 L·atm/mol·K)(273 K)
n ≈ 3.389 moles
Since the mole ratio between CaCO3 and CO2 is 1:1, the number of moles of CaCO3 needed will also be approximately 3.389 moles.
To convert moles to grams, you need to use the molar mass of calcium carbonate (CaCO3), which is 100.09 g/mol.
Mass of CaCO3 = moles x molar mass = 3.389 moles x 100.09 g/mol ≈ 338.93 grams
Therefore, approximately 338.93 grams of calcium carbonate are needed to produce 81.0 liters of carbon dioxide gas at STP.
So the answer is not 80.95 grams, but rather approximately 338.93 grams.