Calcium carbonate reacts with aqueous HCl to give CaCl2andCO2 according to the raction,CaCO3(s)+2HCl(aq)→CaCl2(aq)+CO2(g)+H2O(l).
The mass of CaCO3 is required to react compelety with 25mL of 0.75MHCI is:
A
0.1g
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
0.5g
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
1.5 g
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
0.94g
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
Open in App
Solution
The correct option is D0.94g CaCO3+2HCl(aq)→CaCl2+CO2+H2O n=25×0.751000
Moles of CaCO3required=25×0.751000×12
Mass of CaCO3=25×0.751000×12×100=0.9375g