paige heated 3.00 g mercury (ii) oxide (hgo, 216.59 g/mol) to form mercury (hg, 200.59 g/mol) and oxygen…

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?

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

Explanation:

Step1: Calculate moles of HgO

$n_{HgO}=\frac{m}{M}=\frac{3.00\ g}{216.59\ g/mol}\approx0.0139\ mol$

Step2: Determine theoretical moles of $O_2$

The decomposition reaction of HgO is $2HgO\rightarrow 2Hg + O_2$. From the stoichiometry, for every 2 moles of HgO, 1 mole of $O_2$ is produced. So, $n_{O_2,theo}=0.0139\ mol\times\frac{1}{2}= 0.00695\ mol$

Step3: Calculate theoretical mass of $O_2$

$m_{O_2,theo}=n\times M = 0.00695\ mol\times32.00\ g/mol = 0.2224\ g$

Step4: Calculate percent - yield of $O_2$

$\text{Percent yield}=\frac{m_{O_2,actual}}{m_{O_2,theo}}\times100%=\frac{0.195\ g}{0.2224\ g}\times100%\approx87.7%$

Answer:

87.7%