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.

i think it's c. i'm not sure but i think so.