When heated, calcium carbonate decomposes to yield calcium oxide and carbon dioxide gas via the reaction
CaCO3(s)->CaO(s)+CO2(g)
What is the mass of calcium carbonate needed to produce 53 L of carbon dioxide at STP?
You do this the same way as the butane/CO2 problem above.
2.3
The decomposition of 3.32 g CaCO3 results in an actual yield of 1.24 g of CaO. What is the percent yield of calcium oxide for this reaction?
CaCO3(s) → CaO(s) + CO2(g)
To determine the mass of calcium carbonate needed to produce 53 L of carbon dioxide gas at STP (Standard Temperature and Pressure), we need to use the molar volume of a gas at STP and the balanced chemical equation.
Step 1: Determine the balanced chemical equation.
The balanced chemical equation for the decomposition of calcium carbonate to yield calcium oxide and carbon dioxide gas is:
CaCO3(s) -> CaO(s) + CO2(g)
Step 2: Find the molar volume of carbon dioxide at STP.
At STP, one mole of any gas occupies a volume of 22.4 liters. This is known as the molar volume.
Step 3: Convert the given volume to moles of carbon dioxide.
To convert the given volume of carbon dioxide gas to moles, we use the following formula:
moles = volume (in liters) / molar volume at STP
moles of CO2 = 53 L / 22.4 L/mol = 2.37 mol
Step 4: Use the balanced chemical equation to calculate the moles of calcium carbonate.
From the balanced chemical equation, we know that 1 mole of calcium carbonate produces 1 mole of carbon dioxide gas. Therefore, the moles of calcium carbonate will be equal to the moles of carbon dioxide gas.
moles of CaCO3 = 2.37 mol
Step 5: Calculate the molar mass of calcium carbonate.
The molar mass of calcium carbonate can be found by adding up the atomic masses of its constituent elements: calcium (Ca), carbon (C), and oxygen (O).
Molar mass of CaCO3 = (40.08 g/mol of Ca) + (12.01 g/mol of C) + (3 × 16.00 g/mol of O) = 100.09 g/mol of CaCO3
Step 6: Calculate the mass of calcium carbonate.
The mass of calcium carbonate can be calculated using the following formula:
mass = moles × molar mass
mass of CaCO3 = 2.37 mol × 100.09 g/mol = 237 g
Therefore, 237 grams of calcium carbonate are needed to produce 53 liters of carbon dioxide gas at STP.