Question

The weight of CaCO3 in grams that can be precipitated from 0.44g of CO2 by passing it in lime water is








Please give me a proper solution with good explanation friends. If it is helpful for me I will definitely mark it as BRAINLIEST...​​

Answer 1

Answer:

Equivalent weight = molecular weight of compound ÷ charge on compound.

Molecular weight of CaCO3 (calcium carbonate):

Atomic weight of calcium+ atomic weight of carbon + atomic weight of oxygen.

Atomic weight of Calcium : 40

Atomic weight of Carbon :12

Atomic weight of Oxygen:16

So, 40+12+16(3)= 100

Charge on CaCO3 = ca^2+ + CO3^2-

Charge = 2.

So equivalent weight of CaCO3 = 100÷2

Equivalent weight of CaCO3 =50

limestone, also CaCO3.

so im assuming the acid here is HCl

first of all the ratio of CO2 to CaCO3 is one-one meaning the moles are the same.

using n=m/M find mol of CO2 hence finding the mol of CaCO3 (worked out to be 0.01

now rearrange the formula to make the mass the subject (m=nM)

eventually the answer is 1 gram

1/1.25= 0.8

answer =80%