+1993 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
+154 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)
\(2.50g\;H_2*\frac{1mol\;H_2}{2.016g\;H_2}*\frac{1mol\;CO}{2mol\;H_2}=0.620\;mol\;CO\\ 1.34mol\;CO-0.620mol\;CO=0.72mol\;CO\)
+1993 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?
+1993 where did you get 1.33 mol? jsut before that you had said 1.34 mol of CO
+154 For number 2, the answer is limited to 3 sig-figs because of the 2.50g of H2.
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.
+1993 that would then mean the actual annswer there is 5.41 g H2 right ? well rounded to the hundrths place
+154 yep, mb, sorry for these careless mistakes.
1.34*2*2.016=5.40g H2.
Hydrogen remains the limiting reagent.
+1993 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 ?
+154 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.
\(2.50g\;H_2*\frac{1mol\;H_2}{2.016g\;H_2}*\frac{1mol\;CO}{2mol\;H_2}*\frac{22.4L\;CO}{1mol\;CO}=13.9L\;CO\;consumed\\ 30L\;CO-13.9L\;CO=16.1L\;CO\; remaining\)
+1993 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 ?
+154 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.
\(2.50g\;H_2*\frac{1mol\;H_2}{2.016g\;H_2}*\frac{1mol\;CO}{2mol\;H_2}*\frac{28.01g\;CO}{1mol\;CO}=17.37g\;CO\;consumed\\ 30.0L\;CO*\frac{1mol\;CO}{22.4L\;CO}*\frac{28.01g\;CO}{1mol\;CO}=37.5g\;CO\;originally\\ 37.5g-17.37g=20.1g\;CO\;left\;over\)
+1993 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 
