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# Chemsity help

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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

Feb 16, 2019

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

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Feb 16, 2019
edited by ChowMein  Feb 16, 2019
#2
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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
edited by jjennylove  Feb 16, 2019
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where did you get 1.33 mol? jsut before that you had said 1.34 mol of CO

jjennylove  Feb 16, 2019
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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.

ChowMein  Feb 16, 2019
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and mb, should be 1.34 mol

ChowMein  Feb 16, 2019
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this one should be 4 correct, the one in red ?

jjennylove  Feb 16, 2019
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ChowMein  Feb 16, 2019
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that would then mean the actual annswer there is 5.41 g H2 right ? well rounded to the hundrths place

jjennylove  Feb 16, 2019
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yep, mb, sorry for these careless mistakes.
1.34*2*2.016=5.40g H2.

Hydrogen remains the limiting reagent.

ChowMein  Feb 16, 2019
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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
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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
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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
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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
edited by ChowMein  Feb 16, 2019
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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
edited by jjennylove  Feb 16, 2019
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basically all of them work as they can be converted to each other. it just depends on what the question asks for.

0.72mol, 16.1L, 20.1g all work and can be converted to and from each other

ChowMein  Feb 16, 2019
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Which one is the question asking for ? I have no idea

jjennylove  Feb 20, 2019
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Chow Mein reacts with Lo Mein producing Chop Suey

The limiting agent is Chow Mein

As fucked up as this looks, you need to start over to check if it’s correct.

Feb 16, 2019