Respuesta :
Answer:
(B) 42.1%
Explanation:
The formula for the calculation of moles is shown below:
[tex]moles = \frac{Mass\ taken}{Molar\ mass}[/tex]
Given: For [tex]Na_2S[/tex]
Given mass = 15.5 g
Molar mass of [tex]Na_2S[/tex] = 78.0452 g/mol
Moles of [tex]Na_2S[/tex] = 15.5 g / 78.0452 g/mol = 0.1986 moles
Given: For [tex]CuSO_4[/tex]
Given mass = 12.1 g
Molar mass of [tex]CuSO_4[/tex] = 159.609 g/mol
Moles of [tex]CuSO_4[/tex] = 12.1 g / 159.609 g/mol = 0.0758 moles
According to the given reaction:
[tex]Na_2S+CuSO_4\rightarrow Na_2SO_4+CuS[/tex]
1 mole of [tex]Na_2S[/tex] react with 1 mole of [tex]CuSO_4[/tex]
0.1986 mole of [tex]Na_2S[/tex] react with 0.1986 mole of [tex]CuSO_4[/tex]
Available moles of [tex]CuSO_4[/tex] = 0.0758 moles
Limiting reagent is the one which is present in small amount. Thus, [tex]CuSO_4[/tex] is limiting reagent. (0.0758 < 0.1986)
The formation of the product is governed by the limiting reagent. So,
1 mole of [tex]CuSO_4[/tex] gives 1 mole of [tex]CuS[/tex]
0.0758 mole of [tex]CuSO_4[/tex] gives 0.0758 mole of [tex]CuS[/tex]
Molar mass of [tex]CuS[/tex] = 95.611 g/mol
Mass of [tex]CuS[/tex] = 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.
From the information provided, the percent yield of the reaction is 42.1% .
What is percent yield?
The percent yield is the ratio of the actual yied to the theoretical yield multiplied by 100. We have the reaction; Na2S + CuSO4 → Na2SO4 + CuS
Number of moles of Na2S = 15.5 g /78 g/mol = 0.199 moles
Number of moles of CuSO4 = 12.1 g/160 g/mol = 0.076 moles
Since the reaction is 1:1, CUSO4 is the limiting reactant.
1 mole of CUSO4 yields 1 mole of CuS
0.076 moles of CUSO4 yields 0.076 moles CuS
Mass of CuS = 0.076 moles * 96 g/mol = 7.3 g
Percent yield = 3.05 g/7.3 g * 100 = 42.1%
Learn more about percent yield:https://brainly.com/question/1190311
