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: The percent yield of the reaction is 75 %
Explanation:
We are given:
Moles of HgO decomposed = 4.00 moles
The given chemical reaction follows:
[tex]2HgO(s)\rightarrow 2Hg(l)+O_2(g)[/tex]
By Stoichiometry of the reaction:
2 moles of HgO produces 1 moles of oxygen gas
So, 4.00 moles of HgO will produce = [tex]\frac{1}{2}\times 4.00=2mol[/tex] of oxygen gas
To calculate the percentage yield of the reaction, we use the equation:
[tex]\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100[/tex]
Experimental yield of oxygen gas = 1.50 moles
Theoretical yield of oxygen gas = 2.00 moles
Putting values in above equation, we get:
[tex]\%\text{ yield of oxygen gas}=\frac{1.50}{2.00}\times 100\\\\\% \text{yield of oxygen gas}=75\%[/tex]
Hence, the percent yield of the reaction is 75 %
