Question

Consider the following thermochemical equation: CaCO3(s) CaO(s) + CO2 Ho = +178 kJ

a) How many moles of CaCO3 are in 12.0g of CaCO3? moles

b) How much heat must be absorbed by 12.0 g of CaCO3 to convert it completely to CaO? kJ

Answer 1

a)

Molar mass of CaCO3 = 1*MM(Ca) + 1*MM(C) + 3*MM(O)

= 1*40.08 + 1*12.01 + 3*16.0

= 100.09 g/mol

mass of CaCO3 = 12.0 g

we have below equation to be used:

number of mol of CaCO3,

n = mass of CaCO3/molar mass of CaCO3

=(12.0 g)/(100.09 g/mol)

= 0.12 mol

Answer: 0.12 mol

b)

from reaction,

when 1 mol of CaCO3 reacts, heat absorbed = 178 KJ

So,

for 0.12 mol, heat absorbed = 0.12*178 KJ

= 21.4 KJ

Answer: 21.4 kJ