Question

Sulfur and oxygen react to produce sulfur trioxide. In a particular experiment, 7.9 grams of SO3 are produced by the reaction of 5.0 grams of O2 with 6.0 grams of S. What is the % yield of SO3 in this experiment?

Answer 1

Answer:

$\%\, yield\, \, SO_3=82.29$

Explanation:

First write the balance eqation of chemical reaction:

$2S(s) +3O_2(g) \rightarrow 2SO_3(g)$

Remember writing any chemical reaction from its  elemetal form then write the elements in their natural form i.e. how that element exists in the nature here sulphur exists in solid monoatomic form in the nature and oxygen in gaseous diatomic form.

mass of oxygen given=5gram

mole of oxygen$=5/32mol=0.16mol$

mass of sulpher given=6gram

mole of sulpher$=6/32mol=0.19mol$

from the above balanced equaion;

2 mole of sulphur reacts with 3 mole Oxygen completely

1 mole of sulphur reacts with 1.5 mole Oxygen completely

0.19 mole of sulphur reacts with $1.5\times 0.19$ i.e. 0.285 mole Oxygen completely.

but we have 0.16 mole so oxygen will be the limiting reagent and sulpher will be the excess reagent

so product will depend on the limiting reagent

from the balance equation

3 mole of sulpher gives 2 mole $SO_3$

1 mole of sulpher will give 2/3 mole $SO_3$

0.16 of sulpher will give 0.12 mole $SO_3$

mass of $SO_3$=9.6gram this is theoreical production of $SO_3$

and

actual production of $SO_3$ =7.9gram

$\%\, yield \,\, SO_3=\frac{Actual \,yield}{theoretical\, yield}\times 100$

$\%\, yield\, \, SO_3=\frac{7.9}{9.6} \times 100=82.29$

$\%\, yield\, \, SO_3=82.29$