Question

the mass of CaCO3, produced when carbon dioxide is passed in excess through 500ml of 0.5M Ca(OH)2 will be.......​

Answer 1

Answer:

50 grams is the answer ( the option number C )

Answer 2

Acc. to me,

Your Reaction,

CO2 + Ca(OH)2 ---> CaCo3 +H20

First of all you must know what is given,

For Ca(OH)2

0.5M=Molarity

500ml=0.5L=Volume

Find CaCo3 mass

Steps-----

Step 1: Find ratio in which they react

Step 2: Find real mole

Step 3 : find wt.

Step 1-

Reaction :

CO2 + Ca(OH)2 ---> CaCo3 + H20

 1            1                      1            1      <--------ratio

Step 2 :

mole of Ca(OH)2

Molarity=mole/volume(in L)

0.5=n/0.5

n=1/4

You see all thing are in same ratio so mole are also in same ratio and same magnitude

So mole  of CaCo3 is same as mole of Ca(OH)2

Step 3 -

molecular weight=20 + 12 + 3*16 =100

mole= weight/molecular weight

Given :

mole is same of Ca(OH)2  = 1/4

weight=?

molecular weight =100

On finding wt=25

So answer is D

I hope it Help