Calcium carbonate decomposes at high temperature to form carbon dioxide and calcium oxide:CaCO3 ⟶ CO2 + CaOHow many grams of calcium carbonate are needed to form 93.02 L of carbon dioxide at STP?
Question
Calcium carbonate decomposes at high temperature to form carbon dioxide and calcium oxide:CaCO3 ⟶ CO2 + CaOHow many grams of calcium carbonate are needed to form 93.02 L of carbon dioxide at STP?
Solution 1
To solve this problem, we need to use the ideal gas law and stoichiometry.
Step 1: Use the ideal gas law to find the number of moles of CO2. The ideal gas law is PV = nRT, where P is pressure, V is volume, n is the number of moles, R is the ideal gas constant, and T is temperature. At STP (standard temperature and pressure), the pressure is 1 atm, the volume is 93.02 L (given in the problem), and the temperature is 273.15 K. The ideal gas constant R is 0.0821 L·atm/K·mol. So we can solve for n:
n = PV/RT = (1 atm * 93.02 L) / (0.0821 L·atm/K·mol * 273.15 K) = 4.1 mol CO2
Step 2: Use stoichiometry to find the number of moles of CaCO3. From the balanced chemical equation, we know that 1 mol of CaCO3 produces 1 mol of CO2. So we need 4.1 mol of CaCO3.
Step 3: Convert moles of CaCO3 to grams. The molar mass of CaCO3 is about 100.09 g/mol, so:
4.1 mol CaCO3 * 100.09 g/mol = 410.37 g
So, we need approximately 410.37 grams of calcium carbonate to form 93.02 L of carbon dioxide at STP.
Solution 2
To solve this problem, we need to use the ideal gas law and stoichiometry.
Step 1: Use the ideal gas law to find the number of moles of CO2. The ideal gas law is PV = nRT, where P is pressure, V is volume, n is the number of moles, R is the gas constant, and T is temperature. At STP (standard temperature and pressure), P = 1 atm, V = 93.02 L, R = 0.0821 L.atm/(mol.K), and T = 273.15 K.
n = PV/RT = (1 atm * 93.02 L) / (0.0821 L.atm/(mol.K) * 273.15 K) = 4.15 mol of CO2
Step 2: Use stoichiometry to find the number of moles of CaCO3. From the balanced chemical equation, we know that 1 mol of CaCO3 produces 1 mol of CO2. Therefore, we need 4.15 mol of CaCO3 to produce 4.15 mol of CO2.
Step 3: Convert moles of CaCO3 to grams. The molar mass of CaCO3 is approximately 100.09 g/mol.
grams of CaCO3 = 4.15 mol * 100.09 g/mol = 415.37 g
Therefore, 415.37 g of calcium carbonate are needed to form 93.02 L of carbon dioxide at STP.
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