What is the percent yield of cus for the following reaction given that you start with 15.5 g of na2s and 12.1 g cuso4? the actually amount of cus produced was 3.05 g. reaction: na2s + cuso4 → na2so4 + cus (a) 16.1% (b) 42.1% (c) 18.93% (d) 7.25% (e) not enough information

(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

Moles of = 15.5 g / 78.0452 g/mol = 0.1986 moles

Given: For

Given mass = 12.1 g

Molar mass of = 159.609 g/mol

Moles of = 12.1 g / 159.609 g/mol = 0.0758 moles

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 %

Option B is correct.

c is having 4 valence electrons and is bonded to three hydrogens and one sulphur atom, therefore, the number of bond pair is equal to four. as all the valence electrons are used in forming the bonds so, there is no lone pair left on carbon.

s is having 6 valence electrons and is bonded to one other sulfur atom and one carbon atom, therefore, the number of bond pair is equal to 2. as only two valence electrons are used in forming the bonds so, four electrons are left. therefore, two lone pairs are present on s.

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