100.0 g of copper(II) carbonate (molar mass = 123.56 g/mol) was heated until it decomposed completely. The gas was collected and cooled to room temperature and normal pressure (25¨¬C and 1.00 atm). What volume of carbon dioxide was produced? The reaction is: CuCO3(s) ¡æ CuO(s) + CO2(g)
2 answers
2.310
To calculate the volume of CO2 produced, we first need to find the number of moles of CO2 produced using stoichiometry:
1 mol CuCO3 produces 1 mol CO2
So, the number of moles of CO2 produced can be calculated as:
mol CO2 = mol CuCO3 = 100.0 g / 123.56 g/mol = 0.809 mol
Next, we can use the ideal gas law to calculate the volume of CO2 at room temperature and normal pressure:
PV = nRT
where:
P = pressure = 1.00 atm
V = volume (unknown)
n = number of moles = 0.809 mol
R = gas constant = 0.0821 L·atm/mol·K
T = temperature = 25°C + 273.15 = 298.15 K
Plugging in these values and solving for V, we get:
V = nRT/P = (0.809 mol)(0.0821 L·atm/mol·K)(298.15 K)/(1.00 atm) = 17.54 L
Therefore, 17.54 L of CO2 was produced.
1 mol CuCO3 produces 1 mol CO2
So, the number of moles of CO2 produced can be calculated as:
mol CO2 = mol CuCO3 = 100.0 g / 123.56 g/mol = 0.809 mol
Next, we can use the ideal gas law to calculate the volume of CO2 at room temperature and normal pressure:
PV = nRT
where:
P = pressure = 1.00 atm
V = volume (unknown)
n = number of moles = 0.809 mol
R = gas constant = 0.0821 L·atm/mol·K
T = temperature = 25°C + 273.15 = 298.15 K
Plugging in these values and solving for V, we get:
V = nRT/P = (0.809 mol)(0.0821 L·atm/mol·K)(298.15 K)/(1.00 atm) = 17.54 L
Therefore, 17.54 L of CO2 was produced.
3 answers