Question

If 4 moles of each reactant are available for the reaction described by the following equation, SiO2(s)+3C(s)-> SiC(s)+ 2CO(g)
_________ grams is the theoretical Yield of CO2
(round your answer to 3 Significant figures)

My cousin's high school problem. I forgot how to do it, couldn't help her. Any help is appreciated.

Answer 1

\[SiO_{2} + 3C_{s} \rightarrow SiC_{s} + 2CO_{g}\]

We note that we have the same number of atoms for each compound on both sides so it's balanced.

Answer 2

@moonlight93 see what i've done so far for this.

now we know that we've got

4 moles of SiO2

and

4 moles of Carbon C

our job now is to find the limiting reagent. the limiting reagent is the substance that's going to run out FIRST.

to do this, we need to first multiply each by the molar ratio. THIS is why balancing the equation becomes important

When we do this, this will give us the amount of moles that we need to make the reaction happen in a 1:1 ratio.

\[4~moles~SiO_{2}*(\frac{ 3C }{ SiO_{2} }) = 12~moles~of~carbon\]

\[4~moles~C*(\frac{ SiO_{2}}{ 3C }) = \frac{ 4 }{ 3 }~moles~of~SiO_{2}\]

Now RECAP

we have 4 moles of Sio2
and 4 moles of carbon

we need

12 moles of carbon

and

only

4/3 moles of SiO2


So from this we know that we're going to run out of carbon very fast. so carbon here is our limiting reagent.

CARBON is going to run out first so what we do now is we figure out the mass of CO.

Answer 3

\[4~moles~C*(\frac{ 2CO }{3C }) = \frac{ 8 }{ 3 }~moles~of~CO \]

to figure out the theoretical yield

\[(\frac{ 8 }{ 3 })moles~CO*(molar~mass~CO)\]

Answer 4

I did it, thanks.