Question

@perl

Answer 1

http://prntscr.com/5l0spf

Answer 2

so first we balance the equation

Answer 3

Yep

Answer 4

Start with the carbons first.

Answer 5

\(\sf C_6H_6+O_2\rightarrow\color{red}{6}CO_2+H_2O\)

Answer 6

Good. Now the hydrogens.

Answer 7

\(\sf C_6H_6+O_2\rightarrow\color{red}{6}CO_2+\color{orange}{3}H_2O\)

Answer 8

Now you have carbons balanced hydrogens balanced, and on the right side you have 12 + 3 oxygens on the right side

Answer 9

15 oxgens on the right, and 2 oxygens on the left.

Answer 10

But 2 doesn't go into 15?

Answer 11

put a coefficient of 15/2 in front of O2

Answer 12

then multiply both sides of the equation by 2

Answer 13

Use dimensional analysis. Here is the pathway you will be taking: \(\sf \color{red}{starting~mole~Benzene} \rightarrow n~mole~CO_2\)

Answer 14

But like they said, you need to fully balance this equation first before you proceed.

You must ALWAYS take care of the chemistry before you can do the math.

Answer 15

\(\sf (C_6H_6+\color{blue}{15/2}O_2\rightarrow\color{red}{6}CO_2+\color{orange}{3}H_2O)*2=2C_6H_6+\color{blue}{15}O_2\rightarrow\color{red}{12}CO_2+\color{orange}{6}H_2O\)

Answer 16

looks good :)

Answer 17

There you go, it's balanced

Answer 18

Now you're given 12.8 moles of benzene, so you can do a mole to mole ratio to find your limiting reactant.

Answer 19

now the equation says
there are 2 moles of benzene reactant for every 12 moles of oxygen product

Answer 20

compare what you already HAVE to how many moles of substance you NEED.

Answer 21

Now, you have: \(\sf \color{blue}{12.8~ ~{mol~C_6H_6}\times \frac{12~mol~CO_2}{2~mol~C_6H_6}}\)

Answer 22

And you're done. Now close this.

Answer 23

you didn't have to close it

Answer 24

lol

Answer 25

:)

Answer 26

but yes, abbot is right

Answer 27

76.8?

Answer 28

yes

Answer 29

76.8 moles of CO2

Answer 30

Thanks @perl and @abb0t :)

Answer 31

and @Jhannybean

Answer 32

:( I only have 1 medal

Answer 33

to give

Answer 34

Give it to me.

Answer 35

lol