4.430 mixture containing CaCO_3(s) is heated until all CaCO_3(s) is decomposed c

Question

4.430 mixture containing CaCO_3(s) is heated until all CaCO_3(s) is decomposed completely. CaCO_3(s) rightarrow CaO (s) + CO_2 (g) (molar mass: CaCO_3= 100.0, CaO=56.1, CO_2 = 44.0 g/mole) If the mass of the mixture remaining after heating is 3.580 g, calculate the percent CaCO_3 in the mixture from the mass loss of the sample. The CO_2(g) liberated from the above reaction is collected over water at 27 degree C and a total pressure of 750. torr. The volume of CO_2(g) is 430. ml and the vapor pressure of water at 27 degree C is 26.7 torr. Calculate the percent CaCO_3(s) in the mixture from data on collected gas. (Assume that CO_2(g) does not dissolve in water.) Setup:

Explanation / Answer

CaCO3---------> CaO + CO2

amount of CaCO3 = 4.430g

moles of CaCO3 = 4.430g/100gmol-1 = 0.0443 mol

amount of CaO produced in the reaction : 0.0443 mol (1 mole of CaCO3 gives one mole CaO)

Molar mass of CaO = 56.1 g/mol

Amount of CaO produced = 2.48g g

Amount of CaCO3 left over = 4.43-2.48 = 1.95g

So percentage of CaCO3 decomposed = (1.95/4.430) x 100 = 44%

b) V= 430 mL = 0.43L

T = 273 + 27 = 300K

P = 750-26.7 = 723.3 torr = 723.3 x 0.00131 = 0.947 atm

n= ?

PV = nRT

n = PV/RT = (0.947 atm x 0.43 L)/(0.0821LatmK-1mol-1 x 300K)

n = 0.0165 mol

0.0165 mole of CO2 means 0.0165 moles of CaCO3 has reacted.

So the amount of CaCO3 left over = 0.0443-0.0165 = 0.0278 moles

The amount of CaCO3 in the mixture = (0.0278/0.0443 ) x 100 = 62.75%

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