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If 88.0 g of CO2 is produced from the complete decomposition of pure CaCO3, how much carbonate is decomposed?

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CuCO3(s) ----------------------> CuO (s) + CO2 (g)

First, we have to check if the reaction is balanced.

We have the same numbers of atoms on the left of the reaction and the same numbers of atoms on the right =)

We have 88 g of CO2:

CuCO3(s) ----------------------> CuO (s) + CO2 (g)

x 88.0 g

x is our Unknow value.

We have to calculate the mass per mole of CuCO3:

1mol CuCO3 = 123.5 g,

and from CO2:

1 mol CO2 = 44.0 g

Then,

123.5 g CuCO3 --------- 44.0 g CO2

x --------- 88.0 g CO2

Answer x = 247 g of CuCO3

The question says how much CaCO3 decomposed, so we have to use this reaction.

CaCO3 (s) --------------> CaO(s) + CO2 (g)

x 88.0 g

1 mol CaCO3 = 100.08 g

1 mol CO2 = 44.0 g

100.08 g CaCO3 ---------- 44.0 g CO2

x ---------- 88.0 g CO2

x= 200.16 g CaCO3 descomposed

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