Question

Please help me understand how to...
Calculate the mass of Cr2O3 that can be produced if the reaction of 56.2 g of chromium and sufficient oxygen has a 76.0 % yield.
...Thank you!

Answer 1

Answer:

$m_{Cr_2O_3}^{actual}=62.4gCr_2O_3$

Explanation:

Hello!

In this case, according to the reaction:

$4Cr+3O_2\rightarrow 2Cr_2O_3$

We can see there is a 4:2 mole ratio between chromium and chromium (III) oxide, this, for the given 56.2 g of chromium, the theoretical yield of the oxide product is computed down below:

$m_{Cr_2O_3}^{theoretical}=56.2gCr*\frac{1molCr}{52.0gCr}*\frac{2molCr_2O_3}{4molCr} *\frac{151.99gCr_2O_3}{1molCr_2O_3} =82.13gCr_2O_3$

Now, considering the 76.0-% yield for this reaction, the actual yield turns out:

$m_{Cr_2O_3}^{actual}=82.13gCr_2O_3*\frac{76.0gCr_2O_3}{100gCr_2O_3} \\\\m_{Cr_2O_3}^{actual}=62.4gCr_2O_3$

Best regards!