Question

Red mercury (II) oxide decomposes to form mercury metal and oxygen gas according to the following equation: 2HgO (s) 2Hg (l) + O2 (g). If 4.00 moles of HgO decompose to form 1.50 moles of O2 and 603 g of Hg, what is the percent yield of this reaction?

Answer 1

hey there!:

2HgO (s) =>  2Hg (l) + O2 (g)

2 moles of HgO decompose to form 2 moles of Hg and 1 mole of O2 according to the reaction mentioned in the question.

So 4.00 moles of HgO must give 4 moles of Hg and 2 moles of O2 theoretically.

603 g of Hg = 603 / 200.6 = 3 moles

Percent yield = ( actual yield / theoretical yield) * 100

= ( 3/4) * 100

= 75 %

Hope this helps!

Answer 2

Answer: The percent yield of the reaction is 75 %

Explanation:

We are given:

Moles of HgO decomposed = 4.00 moles

The given chemical reaction follows:

$2HgO(s)\rightarrow 2Hg(l)+O_2(g)$

By Stoichiometry of the reaction:

2 moles of HgO produces 1 moles of oxygen gas

So, 4.00 moles of HgO will produce = $\frac{1}{2}\times 4.00=2mol$ of oxygen gas

To calculate the percentage yield of the reaction, we use the equation:

$\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

Experimental yield of oxygen gas = 1.50 moles

Theoretical yield of oxygen gas = 2.00 moles

Putting values in above equation, we get:

$\%\text{ yield of oxygen gas}=\frac{1.50}{2.00}\times 100\\\\\% \text{yield of oxygen gas}=75\%$

Hence, the percent yield of the reaction is 75 %