Question

The combustion of glucose is represented by the following balanced equation: C6H12O6+6 O2→6 H2O+6 CO2. The reaction uses 1 gram of both C6H12O6 and O2. What is the percent yield if 0.45 g of H2O is produced? a 0.558% b 100% c 0.31% d 80%

Answer 1

Answer : The correct option is, (d) 80 %

Solution : Given,

Mass of $C_6H_{12}O_6$ = 1 g

Mass of $O_2$ = 1 g

Molar mass of $C_6H_{12}O_6$ = 180 g/mole

Molar mass of $O_2$ = 32 g/mole

Molar mass of $H_2O$ = 18 g/mole

First we have to calculate the moles of $C_6H_{12}O_6$ and $O_2$.

$\text{ Moles of }C_6H_{12}O_6=\frac{\text{ Mass of }C_6H_{12}O_6}{\text{ Molar mass of }C_6H_{12}O_6}=\frac{1g}{180g/mole}=0.00555moles$

$\text{ Moles of }O_2=\frac{\text{ Mass of }O_2}{\text{ Molar mass of }O_2}=\frac{1g}{32g/mole}=0.0312moles$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$C_6H_{12}O_6+6O_2\rightarrow 6H_2O+6CO_2$

From the balanced reaction we conclude that

As, 6 mole of $O_2$ react with 1 mole of $C_6H_{12}O_6$

So, 0.0312 moles of $O_2$ react with $\frac{0.0312}{6}=0.0052$ moles of $C_6H_{12}O_6$

From this we conclude that, $C_6H_{12}O_6$ is an excess reagent because the given moles are greater than the required moles and $O_2$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $H_2O$

From the reaction, we conclude that

As, 6 mole of $O_2$ react to give 6 mole of $H_2O$

So, 0.0312 mole of $O_2$ react to give 0.0312 mole of $H_2O$

Now we have to calculate the mass of $H_2O$

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

$\text{ Mass of }H_2O=(0.0312moles)\times (18g/mole)=0.562g$

Theoretical yield of $H_2O$ = 0.562 g

Experimental yield of $H_2O$ = 0.45 g

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

$\% \text{ yield of reaction}=\frac{\text{ Experimental yield of }H_2O}{\text{ Theoretical yield of }H_2O}\times 100$

$\% \text{ yield of reaction}=\frac{0.45g}{0.562g}\times 100=80\%$

Therefore, the percent yield of reaction is, 80 %

Answer 2

Answer:

The yield would D. 80%!

Explanation:

Since 1 gram of O2 only produces 0.56 g of H2O, whereas 1 g of C6H12O6 produces 0.60 g of H2O, the O2 is the limiting reagent.