Answer:
14.5 g silver
Explanation:
This is a problem using the stoichiometry of the reaction. First thing we need is the balanced equation:
Zn + 2 AgNO3 ----------------------- 2 Ag + Zn(NO3)2
We know that 14.6 g of Zn did not reacted, then we can calculate the amount of Zn reacted and do the calculation given the above reaction.
amount Zn reacted: 19.0 -14.6 g Zn = 4.4 g Zn
atomic weight of Zn: 65.37 g/mol
mol Zn reacted: 4.4 g Zn x ( 1 mol Zn/ 65.37 g Zn) = 0.067 mol Zn
We know from the balanced equation that moles of Ag are produced from 1 mol Zn therefore the mol of Ag produced are:
0.067 mol Zn x 2 mol Ag/ 1mol Zn = 0.135 mol Ag
and the mass of silver then will be given by multiplying by the atomic weight of silver:
0.135 mol Ag x 107.9 g/mol = 14.5 g Ag
Here we have to calculate the amount of
ion present in the sample.
In the sample solution 0.122g of
ion is present.
The reaction happens on addition of excess BaCl₂ in a sample solution of potassium sulfate (K₂SO₄) and sodium sulfate [(Na)₂SO₄] can be written as-
K₂SO₄ = 2K⁺ + 
(Na)₂SO₄=2Na⁺ + 
Thus, BaCl₂+
= BaSO₄↓ + 2Cl⁻ .
(Na)₂SO₄ and K₂SO₄ is highly soluble in water and the precipitation or the filtrate is due to the BaSO₄ only. As a precipitation appears due to addition of excess BaCl₂ thus the total amount of
ion is precipitated in this reaction.
The precipitate i.e. barium sulfate (BaSO₄)is formed in the reaction which have the mass 0.298g.
Now the molecular weight of BaSO₄ is 233.3 g/mol.
We know the molecular weight of sulfate ion (
) is 96.06 g/mol. Thus in 1 mole of BaSO₄ 96.06 g of
ion is present.
Or. we may write in 233.3 g of BaSO₄ 96.06 g of
ion is present. So in 1 g of BaSO₄
g of
ion is present.
Or, in 0.298 g of the filtered mass (0.298×0.411)=0.122g of
ion is present.
Answer: Percent recovery is 47.34 %
Explanation:
Percent yield is defined as the ratio of experimental yiled to theoretical yield in terms of percentage.

Putting in the values we get:

Therefore, the percent recovery is 47.34 %
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