CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

What volume of CO2 at STP (atm) will be produced when 1 g of CaCO3 reacts with an excess of dilute HCl?

A
224 ml
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
112 ml
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
56 ml
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
448 ml
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is A 224 ml
CaCO3+2HClCaCl2+CO2+H2O

1 mol of CaCO3 produces 1 mol of CO2=22.4 L of CO2 gas, i.e.
100 g of CaCO3 produces 22400 mL of CO2 gas.
1 g of CaCO3 will produce = 22400100 mL of CO2 gas.

Volume of CO2 gas produced = 224 mL

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
Join BYJU'S Learning Program
CrossIcon