What volume of CO2 at STP (atm) will be produced when 1g of CaCO3 reacts with an excess of dilute HCl?
A
224ml
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
112ml
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
56ml
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
448ml
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution
The correct option is A224ml CaCO3+2HCl→CaCl2+CO2+H2O
1 mol of CaCO3 produces 1 mol of CO2=22.4L of CO2 gas, i.e. 100g of CaCO3 produces 22400mL of CO2 gas. 1g of CaCO3 will produce = 22400100mL of CO2 gas.