Step-by-step explanation:
The aluminum carbonate will decompose into aluminum oxide and carbon dioxide according to this reaction:
Al₂(CO₃)₃ ----> Al₂O₃ + 3 CO₂
First we will have to convert the grams of aluminum carbonate into moles of it using its molar mass.
molar mass of Al = 26.98 g/mol
molar mass of C = 12.01 g/mol
molar mass of O = 16.00 g/mol
molar mass of Al₂(CO₃)₃ = 2 * 26.98 g/mol + 3 * 12.01 g/mol + 9 * 16.00 g/mol
molar mass of Al₂(CO₃)₃ = 233.99 g/mol
moles of Al₂(CO₃)₃ = 3.75 g/(233.99 g/mol)
moles of Al₂(CO₃)₃ = 0.016 moles
Al₂(CO₃)₃ ----> Al₂O₃ + 3 CO₂
According to the coefficients of the reaction 1 mol of Al₂(CO₃)₃ will produce 3 moles of CO₂. Then the molar ratio between them is 1 to 3. We can use that relationship to find the number of moles of CO₂ that will be produced.
moles of CO₂ = 0.016 moles of Al₂(CO₃)₃ * 3 moles of CO₂/(1 mol of Al₂(CO₃)₃)
moles of CO₂ = 0.048 moles
Now that we know the number of moles of CO₂ produced, we can find the volume of it at 35 °C and 1.45 atm. We can use the ideal gas law formula:
P * V = n * R * T
P = 1.45 atm V = ?
n = 0.048 moles
R = 0.082 atm*L/(mol*K)
T = (273.15 + 35) K
T = 308.15 K
Replacing these values and solving for the volume we will get the answer to our problem.
P * V = n * R * T
V = n * R * T/P
V = 0.048 moles * 0.082 atm*L/(mol*K) * 308.15 K/(1.43 atm)
V = 0.848 L
Answer: 0.848 L of carbon dioxide will be produced.