Question

A 115.0-g sample of oxygen was produced by heating 400.0 g of potassium chlorate.2KClO3 Right arrow. 2KCI + 3O2What is the percent yield of oxygen in this chemical reaction?Use Percent yield equals StartFraction actual yield over theoretical yield EndFraction times 100..

Answer 1

Answer:

73.4% is the percent yield

Explanation:

2KClO₃ →  2KCl  + 3O₂

This is a decomposition reaction, where 2 moles of potassium chlorate decompose to 2 moles of potassium chloride and 3 moles of oxygen.

We determine the moles of salt: 400 g . 1. mol /122.5g= 3.26 moles of KClO₃

In the theoretical yield of the reaction we say:

2 moles of potassium chlorate can produce 3 moles of oxygen

Therefore, 3.26 moles of salt, may produce (3.26 . 3) /2 = 4.89 moles of O₂

The mass of produced oxygen is: 4.89 mol . 32 g /1mol = 156.6g

But, we have produced 115 g. Let's determine the percent yield of reaction

Percent yield = (Produced yield/Theoretical yield) . 100

(115g / 156.6g) . 100 = 73.4 %