Answer :
The percent yield of carbon disulfide is 79.6% when 35.8 g of sulfur dioxide produces 11.3 g of carbon disulfide.
To calculate the percent yield, we first need to determine the theoretical yield of carbon disulfide from the given amount of sulfur dioxide
- Let's consider the balanced chemical equation for the production of carbon disulfide:
[tex]3 SO_2(g) + 4 H_2S(g) → CS_2(l) + 4 H_2O(l)[/tex]
First, calculate the molar masses
- SO2: 32.07 (S) + 2 * 16.00 (O) = 64.07 g/mol
- CS2: 12.01 (C) + 2 * 32.07 (S) = 76.15 g/mol
Next, convert the mass of [tex]SO_2[/tex]to moles:
Moles of SO2 = 35.8 g / 64.07 g/mol ≈ 0.5592 mol
Theoretical moles of [tex]CS_2[/tex]= 0.5592 mol [tex]SO_2[/tex]* (1 mol [tex]CS_2[/tex]/ 3 mol [tex]SO_2[/tex]) ≈ 0.1864 mol
- Convert this amount to grams to find the theoretical yield: Theoretical mass of [tex]CS_2[/tex]= 0.1864 mol * 76.15 g/mol ≈ 14.19 g
Finally, calculate the percent yield using the actual yield (11.3 g) and theoretical yield (14.19 g): Percent yield = (Actual yield / Theoretical yield) * 100%[tex]= (11.3 g / 14.19 g) * 100% ≈ 79.6%[/tex]
= 79.63%
