Answer:
3.4dm3
Explanation:
We'll begin by calculating the number of mole of CaCO3 present in 12.8g of CaCO3.
This can be achieved as shown below:
Molar Mass of CaCO3 = 40 + 12 + (16x3) = 40 + 12 + 48 = 100g/mol
Mass of CaCO3 obtained from the question = 12.8g
Number of mole of CaCO3 =?
Number of mole = Mass /Molar Mass
Number of mole of CaCO3 = 12.8/100
Number of mole of CaCO3 = 0.128 mole
The equation for the reaction is given below:
CaCO3 → CaO + CO2
From the equation above,
1 mole of CaCO3 produced 1 mole of CO2.
Therefore, 0.128 mole of CaCO3 will also produce 0.128 mole of CO2.
Now, we can obtain the volume of CO2 produced as follow:
Data obtained from the question include:
T (temperature) = 38°C = 38 + 273 = 311K
P (pressure) = 0.96 atm
n (number of mole of CO2) = 0.128 mole
R (gas constant) = 0.082atm.dm3/Kmol
V (volume of CO2) =?
Using the ideal gas equation PV = nRT, the volume of CO2 produced can be obtained as shown:
PV = nRT
0.96 x V = 0.128 x 0.082 x 311
Divide both side by 0.96
V = (0.128 x 0.082 x 311) /0.96
V = 3.4dm3
Therefore, 3.4dm3 of CO2 are produced.
