Question

100 g caco3 ,is treated with 1 L of 1 N HCL ,what would be the weight of CO2 after completion of reaction
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Answer 1

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CaCo3+2HCI -->Co2+CaCI2+H2O

1 Mole of CaCo3 reacts with 2 mile of HCI to give

one mole of CO2

1N HCI=1M HCI (n factor for HCI)

Molarity= no of moles/ volume in L

1=no of moles/1L

number of moles of HCI=1

Thus, HCI is limiting reagent.

Hence, only 0.5 moles of CaCo3 react with 1 Mole of HCI to give 0.5 moles of CO2.

0.5= weight/44

Weight of CO2 taken =22//

Answer 2

Answer:

$\huge\bold\red{ANSWER}$

CaCO3 + 2HCl -> CO2 + CaCl2 + H2O

It's clear that 1 mole of CaCO3 reacts with 2 miles of HCl to give 1 mole of CO2.

Now, according to question: 100g CaCO3= 1 mol of CaCO3

1N HCl=1M HCl (n factor for HCl is 1)

Molarity=No of moles/Volume in L

1=No. Of moles/1L

No. Of moles of HCl = 1

Thus, HCl is limiting Reagent.

Hence, only 0.5 moles of CaCO3 would react with 1 mole of HCl to give 0.5 mole of CO2.

Now, 0.5Moles of CO2= weight/molecular mass of CO2.

0.5=weight/44

Weight of CO2 taken= 22g