Question

1.5 moles of clo2 react with 72g of O3 according to the reaction 2clo2+2o3 to give cl2o6+2o2, 0.6 moles of cl2o6 is formed. what is the percentage yield of the reaction?​

Answer 1

Answer: The percentage yield of the reaction is 80 %.

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

• For $O_3$:

Given mass of $O_3$ = 72 g

Molar mass of $O_3$ = 48 g/mol

Putting values in above equation, we get:

$\text{Moles of }O_3=\frac{72g}{48g/mol}=1.5mol$

We are given:

Moles of $ClO_2$ = 1.5 moles

As, moles of both the reactants are equal. Thus, the experimental yield of $Cl_2O_6$ can be calculated from any of the reactants.

For the given chemical reaction:

$2ClO_2+2O_3\rightarrow Cl_2O_6+2O_2$

By Stoichiometry of the reaction:

2 moles of ozone is producing 1 mole of $Cl_2O_6$

So, 1.5 moles of ozone will produce = $\frac{1}{2}\times 1.5=0.75mol$ of $Cl_2O_6$

To calculate the percentage yield of $Cl_2O_6$, we use the equation:

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

Experimental yield of $Cl_2O_6$ = 0.6 mol

Theoretical yield of $Cl_2O_6$ = 0.75 mol

Putting values in above equation, we get:

$\%\text{ yield of }Cl_2O_6=\frac{0.6mol}{0.75mol}\times 100\\\\\% \text{yield of }Cl_2O_6=80\%$

Hence, the percentage yield of the reaction is 80 %

Answer 2

Answer:

80%

Explanation:

2 MOLES OF CLO2 GIVES 1 MOLE OF CL2 O6

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