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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)

Respuesta :

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.

[tex]\text{Moles of }KO_2=\frac{\text{Mass of }KO_2}{\text{Molar mass of }KO_2}[/tex]

[tex]\text{Moles of }KO_2=\frac{27.9g}{71.10g/mol}=0.392mol[/tex]

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 [tex]\frac{29.0}{22.4}=1.29[/tex] mole of CO₂ gas.

Now we have to calculate the limiting and excess reagent.

The balanced chemical equation is:

[tex]4KO_2+2CO_2\rightarrow 2K_2CO_3+3O_2[/tex]

From the balanced reaction we conclude that

As, 4 mole of [tex]KO_2[/tex] react with 2 mole of [tex]CO_2[/tex]

So, 0.392 moles of [tex]KO_2[/tex] react with [tex]\frac{2}{4}\times 0.392=0.196[/tex] moles of [tex]CO_2[/tex]

From this we conclude that, [tex]CO_2[/tex] is an excess reagent because the given moles are greater than the required moles and [tex]KO_2[/tex] is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of [tex]K_2CO_3[/tex]

From the reaction, we conclude that

As, 4 mole of [tex]KO_2[/tex] react to give 2 mole of [tex]K_2CO_3[/tex]

So, 0.392 moles of [tex]KO_2[/tex] react to give [tex]\frac{2}{4}\times 0.392=0.196[/tex] moles of [tex]K_2CO_3[/tex]

Now we have to calculate the mass of [tex]K_2CO_3[/tex]

[tex]\text{ Mass of }K_2CO_3=\text{ Moles of }K_2CO_3\times \text{ Molar mass of }K_2CO_3[/tex]

Molar mass of [tex]K_2CO_3[/tex] = 110.98 g/mole

[tex]\text{ Mass of }K_2CO_3=(0.196moles)\times (138.21g/mole)=27.1g[/tex]

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

[tex]\text{Percent yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100[/tex]

Experimental yield = 21.8 g

Theoretical yield = 27.1 g

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

[tex]\text{Percent yield}=\frac{21.8g}{27.1g}\times 100=80.4\%[/tex]

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

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