Question

1. Refer to this equation:

2CaCO3 (s) —> 2CaO (s) + 2CO2 (g)

Enthalpy change = -178.1 kJ/ mol reaction

How many grams of CaCo3 must react in order to liberate 545 kJ of heat?

2. Refer to this equation:

2Al (s) + Fe2O3 (s) —> 2Fe (s) + Al2O3 (s)

Enthalpy change = -847.6 kJ/ mol reaction

How much heat is released if 35.0 g of Al (s) reacts to completion?

Answer 1

1)

from reaction,

2 moles of CaCO3 liberates 178.1 KJ

So,

to liberate 545 KJ, moles of CaCO3 required = 2*(545/178.1)

= 6.12 mol

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

we have below equation to be used:

mass of CaCO3,

m = number of mol * molar mass

= 6.12 mol * 100.09 g/mol

= 613 g

Answer: 613 g

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