Question

What is the percent yield for the reaction below when
705.0 g SO2 and 80.0 g 0₂ produce 586.0 g SO3?
2SO2(g) + O2(g) → 2SO3(g)

Answer 1

The percent yield would be 68.3%

Percent yield

This is given as:

Percent yield = yield/theoretical yield x 100%

From the equation:

$2SO_2(g) + O_2(g) -- > 2SO_3(g)$

The mole ratio of the reactants is 2;1

Mole of 705 g SO2 = 705/64 = 11.02 moles

Mole of 80 g O2 = 80/32 = 2.5 moles

O2 is limiting.

Mole ratio of O2 and SO3 = 1:2

Mole equivalent of SO3 = 2.5 x 2 = 5.0 moles

Mass of 5 moles SO3 = 5 x 80 = 400 grams

Percent yield = 400/586 = 68.3%

More on percent yield can be found here: https://brainly.com/question/17042787

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