Question

Calcium carbonate, CaCO3(s), decomposes upon heating to give CaO(s) and CO2(g). A sample of CaCO3 is decomposed, and the carbon dioxide is collected in a sealed 250 mL flask. After the decomposition is complete, the gas has a pressure of 1.3 atm at a temperature of 33°C. How many moles of CO2 gas were generated? (Ideal gas equation:

Answer 1

Answer:

0.013 mole

Explanation:

Step 1:

Data obtained from the question. This includes:

Volume (V) = 250 mL

Pressure (P) = 1.3 atm

Temperature (T) = 33°C

Number of mole of (n) =?

Step 2:

Conversion to appropriate units.

The volume and the temperature given in the question must be converted to their appropriate unit in order to obtain the answer to the question in the right unit.

For volume:

We must, convert mL to L

1000 mL = 1 L

Therefore, 250 mL = 250/1000 = 0.25 L

For temperature:

We, must convert °C to K.

K = °C + 273

K = 33°C + 273

K = 306K

Step 3:

Determination of the number of mole of CO2.

Applying the ideal gas equation:

PV = nRT

The number of mole of CO2 can be obtained as follow:

V = 0.25 L

P = 1.3 atm

T = 306K

R (gas constant) = 0.082atm.L/Kmol

Number of mole of (n) =?

PV = nRT

1.3 x 0.25 = n x 0.082 x 306

Divide both side by 0.082 x 306

n = (1.3 x 0.25) /(0.082 x 306)

n = 0.013 mole

Answer 2

Answer:

$n_{CO_2}=0.013molCO_2$

Explanation:

Hello,

In this case, the undergoing chemical reaction is:

$CaCO_3(s)\rightarrow CaO(s)+CO_2(g)$

Thus, over the given conditions, we use the ideal gas equation to compute the moles of carbon dioxide that were generated directly, as it is the only gaseous species:

$PV=n_{CO_2}RT\\\\n_{CO_2}=\frac{PV}{RT} =\frac{1.3atm*250mL*\frac{1L}{1000mL} }{0.082\frac{atm*L}{mol*K}*(33+273.15)K} \\\\n_{CO_2}=0.013molCO_2$

Best regards.