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
Share
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
Sign Up to our social questions and Answers Engine to ask questions, answer people’s questions, and connect with other people.
Login to our social questions & Answers Engine to ask questions answer people’s questions & connect with other people.
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:
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//
hope it's help you