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.
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.
Shahzad said:
8 years ago
How can it be possible? Please explain.
Yasir said:
7 years ago
Please explain this clearly.
Yasir said:
7 years ago
I think HNO3 is limiting. Please explain me, if I am wrong.
Salman GECK said:
5 years ago
CaO+2HNO3 -----> Ca(NO3)2 + H2O.
Yaduvanshi said:
5 years ago
@Ankur.
You are right, thanks.
You are right, thanks.
Hafiz hamid said:
3 years ago
How the answer will be 82?
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.
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