If 20 g of CaCO3 is treated with 50 g of HCl solution, then amount of CO2 produced is
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100g
1mol
CaCO
3
(s)
+
73g
2mol
2HCl(aq.)
→CaCl
2
(aq.)+H
2
O(l)+
44g
1mol
CO
2
(g)
Let CaCO
3
(s) be completely consumed in the reaction.
∵100gCaCO
3
give 44gCO
2
∴20gCaCO
3
will give
100
44
×20gCO
2
=8.8gCO
2
Let HCl be completely consumed.
∵ 73 g HCl give 44 g CO
2
∴ 20 gHCl will give
73
44
×20gCO
2
=12.054gCO
2
Since, CaCO
3
gives least amount of product CO
2
, hence CaCO
3
is limiting reactant. Amount of CO
2
formed will 8.8 g.
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