When producing a soluble salt in a reaction between an acid and an alkali, how can you prepare dry solid crystals from the solut
1 answer:
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Answer:
(B) 42.1%
Explanation:
The formula for the calculation of moles is shown below:
Given: For
Given mass = 15.5 g
Molar mass of = 78.0452 g/mol
<u>Moles of = 15.5 g / 78.0452 g/mol = 0.1986 moles</u>
Given: For
Given mass = 12.1 g
Molar mass of = 159.609 g/mol
<u>Moles of = 12.1 g / 159.609 g/mol = 0.0758 moles</u>
According to the given reaction:
1 mole of react with 1 mole of
0.1986 mole of react with 0.1986 mole of
Available moles of = 0.0758 moles
Limiting reagent is the one which is present in small amount. Thus, is limiting reagent. (0.0758 < 0.1986)
The formation of the product is governed by the limiting reagent. So,
1 mole of gives 1 mole of
0.0758 mole of gives 0.0758 mole of
Molar mass of = 95.611 g/mol
Mass of = Moles × Molar mass = 0.0758 × 95.611 g = 7.2473 g
Theoretical yield = 7.2473 g
Given experimental yield = 3.05 g
% yield = (Experimental yield / Theoretical yield) × 100 = (3.05/7.2473) × 100 = 42.1 %
<u>Option B is correct.</u>