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

Explanation:

Step1: Write the balanced chemical equation

$2HgO \rightarrow 2Hg+O_2$

Step2: Calculate the moles of HgO

The molar - mass of HgO is $M_{HgO}=216.59\ g/mol$. The mass of HgO is $m_{HgO} = 3.00\ g$. The number of moles of HgO, $n_{HgO}=\frac{m_{HgO}}{M_{HgO}}=\frac{3.00\ g}{216.59\ g/mol}\approx0.0139\ mol$.

Step3: Calculate the theoretical moles of $O_2$

From the balanced chemical equation, the mole - ratio of HgO to $O_2$ is 2:1. So the theoretical moles of $O_2$, $n_{O_2,theo}=\frac{1}{2}n_{HgO}=\frac{1}{2}\times0.0139\ mol = 0.00695\ mol$.

Step4: Calculate the theoretical mass of $O_2$

The molar - mass of $O_2$ is $M_{O_2}=32.00\ g/mol$. The theoretical mass of $O_2$, $m_{O_2,theo}=n_{O_2,theo}\times M_{O_2}=0.00695\ mol\times32.00\ g/mol = 0.2224\ g$.

Step5: Calculate the percent yield

The percent yield formula is $\text{Percent Yield}=\frac{m_{O_2,actual}}{m_{O_2,theo}}\times100%$. The actual mass of $O_2$ is $m_{O_2,actual}=0.195\ g$. So the percent yield $=\frac{0.195\ g}{0.2224\ g}\times100%\approx87.7%$.

Answer:

87.7%