Question

the mass of co2 that can be prepared by treating 100 grams of limestone with HCl is​

Answer 1

Explanation:

CaCO3+2HCl→CaCl2+H2O+CO2

mCaCO3=10

MrCaCO3=40+12+16⋅3=100

n=mMr

nCaCO3=10100=0.1

The amount of moles should be proportional to the coefficient in the reaction. Since both CaCO3 and CO2 have the same coefficient:

nCO2=0.1

MrCO2=12+16⋅2=44

mCO2=0.1⋅44=4.4 g

Answer 2

Answer:

your answer is Coco3 + 2hcl to from cacl2 +H2o +co2 -1556249.

Explanation:

I hope your answer is write.