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