If 12.8 g of CaCO3 decomposes at 38 degrees C and 0.96 atm, how many dm3 of CO2 are formed in addition to CaO? CaCO3 → CaO + CO2

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.

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