Question

"In a decomposition process, 2.40 g of impure CaCO3 (CaCO3 mixed with other substances that remain as residue) was decomposed at high temperature and a total pressure of 960.0 torr.

2.00 L CO2 gas was collected over water at 25.0 C. The Water Vapor Pressure is 26.0 torr at 25.0 C. Find Percent by mass of the pure CaCO3"

Thank you!

Answer 1

Calculation of number of moles of CO2 :

We know that ideal gas equation is PV = nRT

Where

T = Temperature = 25oC = 25+273 = 298K

P = pressure = total pressure - vapor pressure of water at 25oC = 960.0 - 26.0 = 934 torr x( 1 atm/760 torr)= 1.229 atm

n = No . of moles = ?

R = gas constant = 0.0821 L atm / mol - K

V= Volume of the gas = 2.00L

Plug the values we get n = (PV)/(RT) =0.1005 moles

The balanced reaction is CaCO3 ---> CaO + CO2

1 mole of CO2 produced from 1 mole of CaCO3

So 0.1005 moles of CO2 produced from 0.1005 moles of CaCO3

So mass of CaCO3 present in the sample = number of moles x molar mass

                                                             = 0.1005 mol x 100 g/mol

                                                             = 10.05 g

Percent of CaCO3 = ( mass of CaCO3 / mass of sample ) x100

                            = ( 10.05g / mass of sample ) x100

Here as you given that the mass of sample is 2.40 g is less than that of mass of CaCO3 that was obtained so please check the mass of substance otherwise the volume of the CO2