Scientist A produces 83.67 g KMnO4 while Scientist B produces 81.35 g KMnO4. What is the percent yield for Scientist A?

What is the percent yield for Scientist B?

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The equation for the production of potassium permanganate is as follows:

2 MnO2 + 2 KOH + O2 → 2 KMnO4 + H2

Explanation:

Hello there!

In this case, according to the given chemical equation for the reaction for the production of potassium permanganate, we can see a 2:2 mole ratio of this product to the starting manganese (II) oxide, which means, we can calculate the theoretical yield of the former via stoichiometry:

Now, we are able to compute the percent yields, by using the actual yield each scientist got:

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answer: the value of enthalpy change will be -152.80 kj.

solution:

pressure = 1atm , temperature = = 298 k

given that pressure and temperature are same in both gases , the number of moles of both gases can be determined by the ideal gas equation.

according to reaction 3 moles no are reacting with 1 mol of ,then 0.49 mol of no will react with moles of ozone that is 0.16 moles of ozone. since, no is limiting reagent and is an excessive reagent.

enthalpy change for the reaction between the given volumes of the gases is :