Question

Sulfur and oxygen react in a combination reaction to produce sulfur trioxide, an environmental pollutant: 2S(s) + 3O2(g) → 2SO3(g) In a particular experiment, the reaction of 1.0 g S with 1.0 g O2 produced 0.80 g of SO3. The % yield in this experiment is __________.

Answer 1

Answer:

54\%

Explanation:

Hello,

In this case, the first step is to identify the limiting reagent by considering the shown below procedure in terms of the reacting sulfur:

$n_S^{available}=1.0gS*\frac{1molS}{32gS}=0.03125molS\\ n_S^{consumed\ by\ O_2}=1.0gO_2*\frac{1molO_2}{36gO_2}*\frac{2molS}{3molO_2} =0.0185molS$

Thus, since 1.0 g of oxygen would consume 0.0185 mole of sulfur and there are 0.03125 mole available, oxygen is the limiting reagent, in such a way, the theoretical yield turns out:

$m_{SO_3}^{theoretical}=0.0185molS*\frac{2molSO_3}{2molS}*\frac{80gSO_3}{1molSO_3}=1.48gSO_3$

Finally, the percent yield:

$\% Y=\frac{0.80g}{1.48g}*100\%=54\%$

Best regards.