40g or Caco3, (s) was treated with 48g of HCL. If the acid used was only 30% strength,
the approximate amount of Caco3, unreacted is
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70% of CaCo3 is left unreacted
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Answer:
Ca CO 3
+2HCl→CaCl
2 +H 2 O+CO 2
Molar mass of Ca CO 3=100 g
Molar mass of HCl =36.5 g
From the reaction, 100 g Ca CO 3
reacts with 73 g HCl
Thus 40 g Ca CO 3
will reacts with = 100 73
×40=29.2 g HCl
Given HCl=40 g, therefore Ca CO 3
is completely consumed.
The limiting reagent, HCl is in excess by 40−29.2=10.8g
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