Question

Write the balanced reaction using the fewest whole number coefficients to describe the reaction between gaseous hydrogen and gaseous oxygen. Then use the information to solve the problem below: In one experiment, 14.0 moles of hydrogen and 10 moles of oxygen were reacted to produce 1.33 moles of H2O. Calculate the % yield.

Answer 1

Answer: The percentage yield of water is 9.5 %

Explanation:

We are given:

Moles of hydrogen = 14 moles

Moles of oxygen = 10 moles

The chemical equation for the formation of water from hydrogen and oxygen follows:

$2H_2+O_2\rightarrow 2H_2O$

By Stoichiometry of the reaction:

2 moles of hydrogen gas reacts with 1 mole of oxygen gas

So, 14 moles of hydrogen gas will react with = $\frac{1}{2}\times 14=7mol$ of oxygen gas

As, given amount of oxygen gas is more than the required amount. So, it is considered as an excess reagent

Thus, hydrogen gas is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

2 moles of hydrogen gas produces 2 moles of water

So, 14 moles of hydrogen gas will produce = $\frac{1}{2}\times 14=7mol$ of water

To calculate the percentage yield of water, we use the equation:

$\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

Experimental yield of water = 1.33 moles

Theoretical yield of water = 14 moles

Putting values in above equation, we get:

$\%\text{ yield of water}=\frac{1.33mol}{14mol}\times 100\\\\\% \text{yield of water}=9.5\%$

Hence, the percent yield of the water is 9.5 %.