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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