Question

Red mercury (II) oxide decomposes to form mercury metal and oxygen gas according to the following equation: 2HgO (s) \(\rightarrow\) 2Hg (l) + O2 (g). If 3.55 moles of HgO decompose to form 1.54 moles of O2 and 618 g of Hg, what is the percent yield of this reaction?

Answer 1

first let's make sure that the equation is balanced

\[2HgO_{s}~\rightarrow 2Hg~+~O_{2}\

make sure that the number of atoms of each element are the same on both the product and reactant side.

do you believe this is balanced and if so why?

Answer 2

FYI this is a decomposition reaction we start with one reactant and get two new products.

Answer 3

@NvidiaIntely once you've told me this we will continue

Answer 4

Okay, well it's not balanced.

Answer 5

let's see

so

we have 2 atoms of Hg and 2 atoms of oxygen on the reactant side.

agree?

Answer 6

No, becuase there on the reactant side it says: 2HgO, and on the product side it says:
2Hg + O2

Answer 7

Which means, that there is 1 oxygen atom on the reactant side, and 2 on the product side.

Answer 8

@NvidiaIntely do you see that 2 in front of HgO?

Answer 9

to find the total atoms you would multiply that by each atom in the compound so

2*HgO would be 2 atoms of Hg and 2 atoms of Oxygen

Answer 10

yeah, you mean the apply's to the gO too?

Answer 11

Okay got yah.

Answer 12

yeah

Answer 13

so our equation is balanced now we get the molar masses of all the compounds


HgO = 216.59 g/mol
Hg = 200.59 g/mol
O2 = 32 g/mol

Answer 14

you understand how to find molar masses right?

Answer 15

Not fully, I would like you to explain further if you have time.

Answer 16

please watch this 3 minute video on how to find the molar mass of a compound

https://www.youtube.com/watch?v=F9NkYSKJifs

Answer 17

Okay, thanks!

Answer 18

Now how do I find the percent yield of this reaction?

Answer 19

we need to use the molar ratios i.e. the numbers in front of the molecules in the balanced equation.

Answer 20

\[\frac{ Hg }{ HgO }*(3.55~mol~HgO)\]

Answer 21

we would multiply the number of moles of HgO by the molar ratio to find out how many moles of each substance are produced.

Answer 22

Hoe do I find the molar ratio?

Answer 23

from the balanced equation, see the numbers in front of the compounds?

Answer 24

Yes.

Answer 25

watch this 2 minute video on stoichiometry

https://www.youtube.com/watch?v=eM_n1HzYlJo

Answer 26

Okay.

Answer 27

so we're given 3.55 moles of HgO

Answer 28

we need to find the number of moles of each product that are produced

Answer 29

so we need to start with 3.55 moles of HgO

Answer 30

Yeah

Answer 31

say if we wanted to find the number of moles of moles of oxygen that should be produced

Answer 32

we multiply the number of moles of HgO by the molar ratio between HgO and O2 ensuring that O2 is in the numerator.

in our balanced equation, O2 we've got 1 mole of that and 2 moles of HgO

So O_{2} goes in the numerator and HgO in the denominator

\[3.55~moles~HgO*(\frac{ O_{2} }{ 2HgO }) = 1.78~mole~O_{2}\]

Answer 33

Okay

Answer 34

now that's based off of the stiochiometry

Answer 35

now we should get 1.78 moles of O2 but we only produced 1.54 moles

Answer 36

Okay

Answer 37

so to find the % yield it's

\[\frac{ No~moles~you~got~from~reaction }{ No~moles~from~stiochiometry }*100 = percent~~yield\]

Answer 38

Here is the % yield of for oxygen

\[\frac{ 1.54 }{ 1.78}*100 = 88~percent\]

now try the same thing for Hg

Answer 39

Okay, let me think.

Answer 40

13.2%

Answer 41

Show your work

Answer 42

First we multiply the number of moles by the molar ratio of the two

\[3.55~moles~HgO~*\frac{ Hg }{ HgO } = 3.55~moles~Hg\]

Answer 43

Okay

Answer 44

now we convert the grams of Hg produced to moles

\[618~grams~Hg*(\frac{ 1~mole~Hg }{ 200.59~grams } )= 3.0~moles~Hg\]

Answer 45

putting it all together we get the following:

\[\frac{ 3.00 }{ 3.55 }*(100) = 87~percent \]

Answer 46

oh, okay....:/

Answer 47

But on my answer sheet/page whatever, it only shows:

13.2%

42.5%

56.6%

86.5%

Answer 48

I got something like 86.7% i may have rounded up somewhere

Answer 49

So then I should use 86.5% right.

Answer 50

yes

Answer 51

Okay thanks!

Answer 52

Also thanks @Photon336 for making my problem sovling SS go to 25!

Answer 53

haha alright no problem