Question

Consider the chemical equation. 2h2 + o2 2h2o what is the percent yield of h2o if 87.0 g of h2o is produced by combining 95.0 g of o2 and 11.0 g of h2? use .

Answer 1

The correct answer is letter C 88.5% 

Answer 2

Answer : The percent yield of water is, 87.87%

Solution : Given,

Mass of hydrogen = 11 g

Mass of oxygen = 95 g

Molar mass of hydrogen, $H_2$ = 2 g/mole

Molar mass of oxygen, $O_2$ = 32 g/mole

Molar mass of water, $H_2O$ = 18 g/mole

Experimental yield of water = 87 g

First we have to calculate the moles of hydrogen and oxygen.

$\text{ Moles of }H_2=\frac{\text{ Mass of }H_2}{\text{ Molar mass of }H_2}=\frac{11g}{2g/mole}=5.5moles$

$\text{ Moles of }O_2=\frac{\text{ Mass of }O_2}{\text{ Molar mass of }O_2}=\frac{95g}{32g/mole}=2.97moles$

Now we have to calculate the moles of water.

The given balanced reaction is,

$2H_2+O_2\rightarow 2H_2O$

As, 2 moles of hydrogen react with 1 mole of oxygen

So, 5.5 moles of hydrogen reaction with $\frac{5.5}{2}=2.75$ moles of oxygen

That means, oxygen is in excess amount and hydrogen is in limited amount. So, we are dealing with the limiting reagent for the calculation of moles of water.

Now we have to calculate the moles of water form the reaction.

As, 2 moles of hydrogen react to give 2 moles of water

So, 5.5 moles of hydrogen react to give 5.5 moles of water

Now we have to calculate the mass of water.

$\text{Mass of }H_2O=\text{Moles of }H_2O\times \text{Molar mass of }H_2O$

$\text{Mass of }H_2O=(5.5moles)\times (18g/mole)=99g$

The theoretical yield of water = 99 g

Now we have to calculate the percent yield of water.

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

% yield of water = $\frac{87g}{99g}\times 100=87.87\%$

Therefore, the percent yield of water is, 87.87%