Question

Chromium(III) oxide can be prepared by heating chromium(IV) oxide in vacuo at high temperature: 4Cr02 —2Cr2O3 +02 The reaction of 480.1 g of CrO2 yields 402.4 g of Cr203. Calculate the theoretical yield of Cr203 (assuming complete reaction) and its percentage yield. Theoretical yield = Percentage yield = Submit Answer 2 question attempts remaining

Answer 1

Answer: The theoretical yield and percent yield of chromium (III) oxide is 434.72 grams and 92.6 % respectively.

Step-by-step explanation:

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

`\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}` .....(1)

Given mass of
`CrO_2` = 480.1 g

Molar mass of
`CrO_2` = 84 g/mol

Putting values in equation 1, we get:

`\text{Moles of }CrO_2=(480.1g)/(84g/mol)=5.72mol`

For the given chemical equation:

`4CrO_2\rightarrow 2Cr_2O_3+O_2`

By Stoichiometry of the reaction:

4 moles of
`CrO_2` produces 2 moles of chromium (III) oxide

So, 5.72 moles of
`CrO_2` will produce =
`(2)/(4)* 5.72=2.86mol` of chromium (III) oxide

Now, calculating the mass of chromium (III) oxide from equation 1, we get:

Molar mass of chromium (III) oxide = 152 g/mol

Moles of chromium (III) oxide = 2.86 moles

Putting values in equation 1, we get:

`2.86mol=\frac{\text{Mass of chromium (III) oxide}}{152g/mol}\\\\\text{Mass of chromium (III) oxide}=(2.86mol* 152g/mol)=434.72g`

To calculate the percentage yield of chromium (III) oxide, we use the equation:

`\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}* 100`

Experimental yield of chromium (III) oxide = 402.4 g

Theoretical yield of chromium (III) oxide = 434.72 g

Putting values in above equation, we get:

`\%\text{ yield of chromium (III) oxide}=(402.4g)/(434.72g)* 100\\\\\% \text{yield of chromium (III) oxide}=\%`

Hence, the theoretical yield and percent yield of chromium (III) oxide is 434.72 grams and 92.6 % respectively.