Question

How many grams of calcium carbonate must be decomposed to produce 5.00 L of carbon dioxide at 901 mmHg and 0'C?

Answer 1

26.3 grams of calcium carbonate must be decomposed to produce 5.00 L of carbon dioxide at 901 mmHg and
`0^0C`

Step-by-step explanation:

According to ideal gas equation:

`PV=nRT`

P = pressure of gas = 901 mm Hg = 1.18 atm (760 mm Hg = 1 atm)

V = Volume of gas = 5.00 L

n = number of moles = ?

R = gas constant =
`0.0821Latm/Kmol`

T =temperature =
`0^0C=(0+273)K=273K`

`n=(PV)/(RT)`

`n=(1.18atm* 5.00L)/(0.0821L atm/K mol* 273K)=0.263moles`

The chemical reaction for the decomposition of calcium carbonate follows the equation:

`CaCO_3(s)\rightleftharpoons CaO(s)+CO_2(g)`

According to stoichiometry:

1 mole of carbon dioxide is produced by = 1 mole of calcium carbonate

Thus 0.263 moles of carbon dioxide is produced by =
`(1)/(1)* 0.263=0.263` moles of calcium carbonate

Mass of calcium carbonate=
`moles* {\text {Molar mass}}=0.263mol* 100g/mol=26.3g`

26.3 grams of calcium carbonate must be decomposed to produce 5.00 L of carbon dioxide at 901 mmHg and
`0^0C`