Question

Dissolving 5.28 g of an impure sample of calcium carbonate in hydrochloric acid produced 1.14 l of carbon dioxide at 20.0 °c and 791 mmhg. Calculate the percent by mass of calcium carbonate in the sample.

Answer 1

Percentage by mass of calcium carbonate in the sample is 93.58%.

Step-by-step explanation:

Assumptions:

• calcium carbonate was dissolved completely and the amount of carbon dioxide released was proportional to the amount of calcium carbonate.
• the sample did not contain any other compound that released or reacted with carbon dioxide.

`PV = nRT`

`n = (PV)/(RT)`

`P = 791 mmHg = (791 mmHg)/(760 mmHg) x 1 atm = 1.040789 atm.`

`R = 8.314 (J)/(K*mol) = 0.082 (L*atm)/(K*mol)`

`T = 20^(0)C + 273^(0) C = 293 K`

`n = (1.040789 atm * 1.14 L)/(0.082 (L * atm)/(K * mol) * 293 K )`

`n = 0.0493839 mol.`

given the equation of the reaction:

`CaCO_(3)  + 2HCL =  CaCl_(2)  + CO_(2) + H_(2)O`

from assumption:

amount carbon dioxide = amount of calcium carbonate

`n(CaCO_(3) ) = n(CO_(2) )`

reacting mass ( m ) = Molar Mass ( M ) * Amount ( n )

`m(CaCO_(3) )   =   n(CaCO_(3) )  *   M(CaCO_(3) )`

m = 0.04938397
`mol` * 100
`(g)/(mol)` = 4.93839
`g`

percentage by mass of CaC
`O_(3)` =
`(mass of pure  )/(mass of impure)*100  = (4.93839g)/(5.28g)*100  = 93.5802%`.