Chemistry, asked by Rockirst, 1 year ago

If 20g of CaCo3 is treated with 20g of HCl how many grams of CO2 will be produced

Answers

Answered by Ashwin0512
27
Here the moles of CaCO3 = Mass of CaCO3/
Molecular Mass
n= 20/100
n= 0.2

Moles of HCl = 20/36.5 = 0.54 mol

The reaction is
CaCO3 + 2HCl = CaCl2 + CO2 + H2O
For 2 mol of HCl, 1 mol CaCO3 is req.

This means for 0.54 mol HCl, 0.27 mol CaCO3 would be req.

But only 0.2 mol CaCO3 is present.
This means CaCO3 is Limiting Reagent.

For 1 mol CaCO3, 1 mol CO2 is formed.

Therefore moles of CO2 produced = 0.27
Mass of CO2 = 44 × 0.27
= 11.88 gm

Rockirst: but how you come to know for 0. 54mol of HCl 0. 27 mol of CaCo3 is req. Only this doubt please clear
Rockirst: But mass of CO2 [answer is 8. 80g in book
Ashwin0512: As we can see from the reaction, for 2 moles of HCl , 1 mol CaCO3 is required . Therefore for 1 mol HCl, 0.5 mol CaCO3 will be required. It means for 0.54 mol HCl, 0.5 × 0.54 = 0.27 mol of CaCO3 will be required.
Rockirst: ok now I understand but Co2 produced is 8. 80 g in Book
Ashwin0512: Sorry, in the last step, I have by mistake said that moles of CO2 will be 0.27 but it should be 0.2 as moles of CaCO3 were 0.2 only. Tjen the answer would be corrected
Ashwin0512: I hope u have understood now
Rockirst: thanks
Answered by write2vaichu
0

Answer:

Explanation:

Here the moles of CaCO3 = Mass of CaCO3/

Molecular Mass

n= 20/100

n= 0.2

Moles of HCl = 20/36.5 = 0.54 mol

The reaction is

CaCO3 + 2HCl = CaCl2 + CO2 + H2O

For 2 mol of HCl, 1 mol CaCO3 is req.

This means for 0.54 mol HCl, 0.27 mol CaCO3 would be req.

But only 0.2 mol CaCO3 is present.

This means CaCO3 is Limiting Reagent.

For 1 mol CaCO3, 1 mol CO2 is formed.

Therefore moles of CO2 produced = 0.27

Mass of CO2 = 44 × 0.27

= 11.88 gm

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