CO(g) + 2 H2 --> CH3OH

2.50 g of hydrogen is reacted with 30.0 L of carbon monoxide at STP.

1. What is the limiting reactant? I think hydrogen

2. What mass of CH3OH is produced? unsure

3. How much excess is left over? unsure

jjennylove Feb 16, 2019

#1**+2 **

1)

\(\text{Find amounts of each reactant in moles}\\ 2.50g\;H_2*\frac{1mol\;H_2}{2.016g\;H_2}=1.24\;mol\;H_2\\ 30.0L\;CO_2*\frac{1mol\;CO}{22.4L\;CO}=1.34\;mol\;CO\\ \text{Find amount of H}_2\text{ that fully reacts with CO or vice versa}\\ 1.34mol\;CO*\frac{2mol\;H_2}{1mol\;CO}*\frac{2.016g\;H_2}{1mol\;H_2}=5.36\;g\;H_2\\ \text{5.36g of hydrogen is needed to fully react with 30.0L of CO, making hydrogen the limiting reagent} \)

2)

\(2.50g\;H_2*\frac{1mol\;H_2}{2.016g\;H_2}*\frac{1mol\;CH_3OH}{2mol\;H_2}*\frac{32.042g\;CH_3OH}{1mol\;CH_3OH}=19.9g\;CH_3OH\)

3)

\(2.50g\;H_2*\frac{1mol\;H_2}{2.016g\;H_2}*\frac{1mol\;CO}{2mol\;H_2}=0.620\;mol\;CO\\ 1.34mol\;CO-0.620mol\;CO=0.72mol\;CO\)

ChowMein Feb 16, 2019

#2**+2 **

I am trying to figure this all out, if you can stay with me a bit longer online. I have psoted anotehr question as well if you can take a look at it.

For this one, wouldnt we round to the hundrths place for the molar masses?

jjennylove
Feb 16, 2019

#3

#4**+1 **

For number 2, the answer is limited to 3 sig-figs because of the 2.50g of H_{2}.

For number 3, the answer is limited to the hundredths place due to the 1.34 mol of CO.

You usually don't want to be limited sig-fig wise when it comes to the molar masses. So, it's best to have the most precise molar mass possible.

ChowMein
Feb 16, 2019

#8**+1 **

that would then mean the actual annswer there is 5.41 g H2 right ? well rounded to the hundrths place

jjennylove
Feb 16, 2019

#9**+1 **

yep, mb, sorry for these careless mistakes.

1.34*2*2.016=5.40g H2.

Hydrogen remains the limiting reagent.

ChowMein
Feb 16, 2019

#10**0 **

its okay, also for 3, why arent we substracting 30.0 L C(amount started with) - 1.24(amount used) = ___

to find the value , why the way you have it set up ?

jjennylove
Feb 16, 2019

#11**+1 **

You can extend the dimensional anaysis-ing to find how much CO is left over in liters as well. This will be equal to the amount of moles of CO left over.

\(2.50g\;H_2*\frac{1mol\;H_2}{2.016g\;H_2}*\frac{1mol\;CO}{2mol\;H_2}*\frac{22.4L\;CO}{1mol\;CO}=13.9L\;CO\;consumed\\ 30L\;CO-13.9L\;CO=16.1L\;CO\; remaining\)

ChowMein
Feb 16, 2019

#12**0 **

but since the question isnt asking for it, we wouldnt do that ? because my questions asks for it in grams ? or could you convert 16.1 into grams ? Which is the correct way ?

jjennylove
Feb 16, 2019

#13**+1 **

If its asking for it in grams, then we can extend the dimensional anaysis-ing to find how much CO is left over in grams. This is also equal to the amount of moles/liters of CO left over. They should all be correct as they are all equivalent amounts.

\(2.50g\;H_2*\frac{1mol\;H_2}{2.016g\;H_2}*\frac{1mol\;CO}{2mol\;H_2}*\frac{28.01g\;CO}{1mol\;CO}=17.37g\;CO\;consumed\\ 30.0L\;CO*\frac{1mol\;CO}{22.4L\;CO}*\frac{28.01g\;CO}{1mol\;CO}=37.5g\;CO\;originally\\ 37.5g-17.37g=20.1g\;CO\;left\;over\)

ChowMein
Feb 16, 2019

#14**0 **

i am sorry, I am getting very confused now there are so many values including the 0.72 and 16.1 L and now these three values . I am unsure which is the correct answer here

jjennylove
Feb 16, 2019