Question

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


please tell me the solution please​

Answer 1

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100g of CaCO3 would be 1 mol because the molecular weight is 100.089. 1 L of 1M HCL would also be 1 mol of HCl, so in the reaction we can write CaCO3 + 2HCl --> CO2 + H2O+ CaCl2

so 1/2 mol of CO2 would come out of the reaction, which would be 22.005 grams because tahts the molecular weigt of CO2

Answer 2

Answer:

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//

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