CaF2 + (NH4)2O --> 2 NH4F + CaO How many grams of NH4F can be formed from 34.6 grams of CaF2? (AKS 4f)
1 answer:
Answer:
32.83 grams of NH₄F can be formed from 34.6 grams of CaF₂
Explanation:
The balanced reaction is:
CaF₂ + (NH₄)₂O ⇒ 2 NH₄F + CaO
By stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction), the following quantities participate in the reaction:
- CaF₂: 1 mole
- (NH₄)₂O: 1 mole
- NH₄F: 2 mole
- CaO: 1 mole
Being the molar mass of the compounds:
- CaF₂: 78 g/mole
- (NH₄)₂O: 52 g/mole
- NH₄F: 37 g/mole
- CaO: 56 g/mole
Then by stoichiometry, the following amounts of mass participate in the reaction:
- CaF₂: 1 mole* 78 g/mole= 78 g
- (NH₄)₂O: 1 mole* 52 g/mole= 52 g
- NH₄F: 2 mole* 37 g/mole= 74 g
- CaO: 1 mole* 56 g/mole= 56 g
You can apply the following rule of three: if 78 grams of CaF₂ form 74 grams of NH₄F by stoichiometry, 34.6 grams of CaF₂ will form how much mass of NH₄F?
mass of NH₄F= 32.83 grams
<u><em>32.83 grams of NH₄F can be formed from 34.6 grams of CaF₂</em></u>
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Explanation:
It is known that of
=
.
(a) Relation between and
is as follows.
Putting the values into the above formula as follows.
=
= 3.347
Also, relation between pH and is as follows.
pH =
=
= 3.44
Therefore, pH of the buffer is 3.44.
(b) No. of moles of HCl added =
=
= 0.0116 mol
In the given reaction, will react with
to form
Hence, before the reaction:
No. of moles of =
= 0.15 mol
And, no. of moles of =
= 0.12 mol
On the other hand, after the reaction :
No. of moles of = moles present initially - moles added
= (0.15 - 0.0116) mol
= 0.1384 mol
Moles of = moles present initially + moles added
= (0.12 + 0.0116) mol
= 0.1316 mol
As, =
=
= 3.347
Since, volume is both in numerator and denominator, we can use mol instead of concentration.
As, pH =
= 3.347+ log {0.1384/0.1316}
= 3.369
= 3.37 (approx)
Thus, we can conclude that pH after the addition of 1.00 mL of 11.6 M HCl to 1.00 L of the buffer solution is 3.37.