elements are arranged in groups by similar atomic structure on the periodic table. this allows for an elements properties to be
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In a chemical reaction, 28g of iron reacts with 16g of sulfur to produce _________ of iron sulfide.
We have the following data:
m(Fe) - mass of iron = 28 g
m(S) - mass of sufur = 16 g
MM(Fe) - molar mass of iron ≈ 56 g/mol
MM(S) - molar mass of sulfur ≈ 32 g/mol
n(Fe) - number of mol of iron = ?
n(S) - number of mol of sulfur = ?
Solving:
* to n(Fe)
* to n(S)
The stoichiometric reaction will be in the same proportion (1 : 1), let us see:
1 mol of Fe -------------- 1 mol of FeS
0.5 mol of Fe ------------ 0.5 mol of FeS
Will the reaction of the iron mass with the mass of sulfur produce how many grams of iron sulfide? We will see:
n(FeS) - number of mol of iron sulfide = 0.5 mol
m(FeS) - mass of iron sulfide = ? (in grams)
MM(FeS) - Molar Mass of iron sulfide = 56 + 32 = 88 g/mol
Solving:
Answer:
44 grams of iron sulfide
___________________________
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.