Chemical Engineering - Stoichiometry - Discussion
Discussion Forum : Stoichiometry - Section 5 (Q.No. 24)
24.
On mixing 56 gm of CaO with 63 gm of HNO3, the amount of Ca(NO3)2 formed is __________ gm.
Discussion:
8 comments Page 1 of 1.
TANMAY SHAH said:
2 years ago
@All,
Hear HNO3 is considered as a limiting reactant.
And as per the formula, the equation we get 2 mol of HNO3 will create 1 mole of Ca(NO3)2.
And HNO3 here given 63 gms.
Now Moles of HNO3 = Mass/ M.W.
No. of Moles = 63/63 = 1 mole
So, 2 mole HNO3 = 1 mole product.
1 mole HNO3 = ?
1/2 moles of product.
Now M.W. of Ca(NO3)2 = 164.
Therefore, Mass = 1/2 * 164 = 82 gm.
Hear HNO3 is considered as a limiting reactant.
And as per the formula, the equation we get 2 mol of HNO3 will create 1 mole of Ca(NO3)2.
And HNO3 here given 63 gms.
Now Moles of HNO3 = Mass/ M.W.
No. of Moles = 63/63 = 1 mole
So, 2 mole HNO3 = 1 mole product.
1 mole HNO3 = ?
1/2 moles of product.
Now M.W. of Ca(NO3)2 = 164.
Therefore, Mass = 1/2 * 164 = 82 gm.
Hafiz hamid said:
3 years ago
How the answer will be 82?
Yaduvanshi said:
5 years ago
@Ankur.
You are right, thanks.
You are right, thanks.
Salman GECK said:
5 years ago
CaO+2HNO3 -----> Ca(NO3)2 + H2O.
Yasir said:
7 years ago
I think HNO3 is limiting. Please explain me, if I am wrong.
Yasir said:
7 years ago
Please explain this clearly.
Shahzad said:
8 years ago
How can it be possible? Please explain.
Ankur said:
10 years ago
If we take Cao as limiting reactant then the answer will be 164 and if we take HNO3 as limiting reactant then the answer will be 82.
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