Question

(A) How to calculate the theoretical yield of water resulting from the complete combustion of \[1.60g\] of methane b) When the reaction was run in the lab \[3.30g\] of water were formed. What is the percent yield of water for this run? \[C{H_4} + 2{O_2} \to C{O_2} + 2{H_2}O\] ?

Answer 1

Hint: First of all we need to start with the balanced chemical equation here where combustion of methane is taking place. Here in the given question we can see \[1:2\] mole ratio between methane and water. So , it determines that the number of moles of water produced should be twice the number of moles of methane which is reacted . Here percent yield can be calculated by using theoretical yield and experimental yield and vice versa we can find each one using this.

Formula used: \[\% \] yield \[ = \] (theoretical yield \[/\] experimental yield ) \[ \times 100\]

Complete step-by-step answer:
First of all let us write the balanced chemical reaction of combustion of methane given as follows, \[C{H_4} + 2{O_2} \to C{O_2} + 2{H_2}O\]
We know that here is \[1:2\] mole ratio between methane and water, therefore ,
Number of moles of produced water \[ \times 2 = \] Number of moles of methane in reaction.

Next, we need to find the theoretical yield of water here.Actually here we should know that the theoretical yield is the yield we get for \[100\% \] reaction when all the methane reacted to form water, therefore the theoretical yield of water thus formed is given by as follows,
\[1.60gC{H_4} \times \dfrac{{1moleC{H_4}}}{{16g}} \times \dfrac{{2mole{H_2}O}}{{1moleC{H_4}}} \times \dfrac{{18g}}{{1mole{H_2}O}}\] \[ = 3.60g\]water
Therefore, theoretical yield of water or other molecules can be calculated in this way and here the theoretical yield of water is \[3.60g\].

Next we have to find the percent yield of water here.As we know the experimental value of water is \[3.30g\] as given in the question. So this means that not all the methane is not reacted here as oxygen acts as the limiting reagent.

The reaction percent yield of water is given by as follows,
\[\% \] yield \[ = \] ( experimental \[/\] actual ) yield \[ \times 100\]
\[\dfrac{{3.30g}}{{3.60g}} \times 100 = 91.7\% \]

Therefore percent yield of water for the run is \[91.7\% \]

Note: First of all from this question we can realize that stoichiometry is an important factor here as we determine the yields based on the stoichiometry as given in the question. Actually we find out the ideal or theoretical yield in accordance with stoichiometry. From the calculated theoretical yield, it helps us to find out the percent yield. Should be familiar with those particular equations. We have to know that in general , due to certain factors which are ranging from purity of chemicals which we used to maintain the perfect reaction condition , the yield of the product we get will always be less than desired value.