Question

Paige heated 3.00 g mercury (II) oxide (HgO, 216.59 g/mol) to form mercury (Hg, 200.59 g/mol) and oxygen (O2, 32.00 g/mol). She collected 0.195 g oxygen. What was the percent yield of oxygen

Answer 1

Answer : The percent yield of oxygen was, 87.84 %

Explanation :

First we have to calculate the moles of $HgO$.

$\text{Moles of }HgO=\frac{\text{Mass of }HgO}{\text{Molar mass of }HgO}=\frac{3g}{216.59g/mole}=0.0139moles$

Now we have to calculate the moles of HgO.

The balanced chemical reaction is,

$2HgO\rightarrow 2Hg+O_2$

From the balanced reaction we conclude that

As, 2 moles of $HgO$ react to give 1 mole of $O_2$

So, 0.0139 moles of $HgO$ react to give $\frac{0.0139}{2}=0.00695$ moles of $O_2$

Now we have to calculate the mass of $O_2$.

$\text{Mass of }O_2=\text{Moles of }O_2\times \text{Molar mass of }O_2$

$\text{Mass of }O_2=(0.00695mole)\times (32g/mole)=0.222g$

The theoretical yield of $O_2$ = 0.222 g

Now we have to calculate the percent yield of $O_2$.

$\%\text{ yield of }O_2=\frac{\text{Actual yield of }O_2}{\text{Theoretical yield of }O_2}\times 100=\frac{0.195g}{0.222g}\times 100=87.84\%$

Therefore, the percent yield of $O_2$ was, 87.84 %

Answer 2

Answer:

87.8 %

Explanation:

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