Determine the theoretical yield and the percent yield if 21.8 g of KCO is produced from reacting 27.9 g KO with 29.0 L of CO (at STP). The molar mass of KO = 71.10 g/mol and KCO = 138.21 g/mol. 4 KO(s) + 2 CO(g) → 2 KCO(s) + 3 O(g)

Answer : The theoretical yield and the percent yield is, 27.1 grams and 80.4 % respectively.

Explanation:

First we have to calculate the moles of KO₂ and CO₂.

and,

As we know that, 1 mole of substance occupies 22.4 L volume of gas.

As, 22.4 L volume of CO₂ present in 1 mole of CO₂ gas

So, 29.0 L volume of CO₂ present in mole of CO₂ gas.

Now we have to calculate the limiting and excess reagent.

The balanced chemical equation is:

From the balanced reaction we conclude that

As, 4 mole of react with 2 mole of

So, 0.392 moles of react with moles of

From this we conclude that, is an excess reagent because the given moles are greater than the required moles and is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of

From the reaction, we conclude that

As, 4 mole of react to give 2 mole of

So, 0.392 moles of react to give moles of

Now we have to calculate the mass of

Molar mass of = 110.98 g/mole

Now we have to calculate the percent yield of the reaction.

Experimental yield = 21.8 g

Theoretical yield = 27.1 g

Now put all the given values in this formula, we get:

Therefore, the percent yield of the reaction is, 80.4 %

yes

explanation:

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