Question

CaCO3 + 2CH3COOH -------> Ca(CH3COO) + H2O + CO2

a) How many moles of CO2 would be required to fill a volume of 11,100L, assuming an atmospheric pressure of 1 atm and a temperature of 77F (298K)?

b) How many moles of acetic acid would be required to generate this much CO2?

Answer 1

Sol :-

a). From ideal gas equation :

PV = nRT ..........................(1)

P = Pressure = 1 atm

V = Volume = 11100 L

n = Number of moles of CO2 gas

R = Universal gas constant = 0.0821 L atm K-1mol-1

and

T = Temperature = 298 K

Substitute all these values in equation (1) :

n = PV/RT

= (1 atm).(11100 L) / (0.0821 L atm K-1mol-1).(298 K)

= 453.69 mol

Hence, number of moles of CO2 gas = 453.69 mol

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b). From the balance chemical equation :

CaCO3 + 2CH3COOH -------> Ca(CH3COO)2 + H2O + CO2

Because,

For 1 mole of CO2 , number of moles of CH3COOH required = 2 mol

So,

For 453.69 mol of CO2 , number of moles of CH3COOH required = 2 mol x 453.69 mol / 1 mol

= 907.38 mol

Hence, number of moles of CO2 required = 907.38 mol