Question

If the reaction of 6.5 grams of C6H12O6 produces 2.5 grams of CO2, what is the percent yield?
C6H12O6 ------> 2C2H5OH + 2CO2

Answer 1

Here is your answer

Hope this will help you

Answer 2

Answer:

The percentage yield of $CO_{2}$ produced observed is  $78.86\%$.

Explanation:

Given,

The mass of $C_{6} H_{12} O_{6}$ = $6.5g$

The mass of $CO_{2}$ produced = $2.5g$

The percentage yield =?

As given, the reaction:

• $C_{6} H_{12} O_{6} \rightarrow C_{2}H_{5}OH + 2CO_{2}$  

As we know,

• The molar mass of $C_{6} H_{12} O_{6}$ = $180g/mol$
• The molar mass of $2$$CO_{2}$ = $2\times 44g/mol$ = $88g/mol$

Now,

• $1 mole$ of $C_{6} H_{12} O_{6}$ yield = $2moles$ of $CO_{2}$

Also,

• $180g$ of $C_{6} H_{12} O_{6}$ yield = $88g$ of $CO_{2}$
• $1g$ of $C_{6} H_{12} O_{6}$ yield = $\frac{88}{180}g$ of $CO_{2}$

Thus,

• $6.5g$ of $C_{6} H_{12} O_{6}$ yield = $6.5\times\frac{88}{180}g$ of $CO_{2}$ = $3.17g$ of $CO_{2}$

Therefore, the theoretical yield calculated = $3.17g$

Now, the percentage yield = $\frac{Experimental yield}{Theoritical yield}\times 100$

Thus, the percentage yield of $CO_{2}$ =  $\frac{2.5}{3.17} \times 100$ = $78.86\%$.