eli123
can someone please help me find these values?
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eli123
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eli123
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@phi
eli123
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@robtobey
NoelGreco
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For someone to comment, he's have to know more about the lab.
aaronq
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the moles you can simply find by dividing the masses by their molar mass... for the volume of H2 formed, you need to write the equation of the reaction in your exp and work with that.
eli123
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@aaronq so for magnesium it would be 24.31/0.035?
aaronq
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the other way around
eli123
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oh ok and for h2? @aaronq
aaronq
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you have to find the volume first
eli123
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is the volume of hydrogen gas produced the same as teh volume of liquid displaced? @aaronq
aaronq
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unless you made other gases in your experiment
aaronq
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yes
eli123
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no i didnt
eli123
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so the volume divided by the mass of h2 will give me the moles of h2?
aaronq
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well for ideal gases you can just use the fact that 1 mole = 22.414 L at STP
aaronq
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otherwise you have to know the density of the gas
eli123
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therefore, how would the equation look?
aaronq
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so for ideal gases,
moles = volume of gas/(22.414 L/mol)
aaronq
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make sure you have the same units
eli123
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oooh
eli123
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@aaronq for the moles of magnesium igot 0.001 for all the trials, but my friend wrote it in dimensional analysis . whats betteR?
aaronq
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it's the same thing, since it's a gas, the volume is proportional to the amount of moles
eli123
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ok and how would i find the ratio of volume displaced in mL to moles h2 gas produced? @aaronq
aaronq
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a ratio is just a division
volume of gas/moles of gas
eli123
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so volume of h2/moles of h2?
eli123
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@aaronq
aaronq
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yeah
eli123
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@aaronq i get a huge number
eli123
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37.00ml/0.002
aaronq
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yep, thats what the want
eli123
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18500?
aaronq
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they actually want the volume displaced/molesof H2 which is the same thing
aaronq
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mL/mol
eli123
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so its right?
eli123
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@aaronq
aaronq
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to what they're asking you, yes
eli123
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but its such a big number? @aaronq
aaronq
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yep thats not that big of a number, remember it's in mL ..
eli123
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18.5 L DAMM
aaronq
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lol that's not a lot, it really should be around 22.414 L
eli123
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ARE YOU POSITIVE?
aaronq
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well unless they want you to calculate using
Mg (s) + HCl -> H2 + MgCl (aq)
eli123
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how would that be?
aaronq
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this would be the theoretical way
you balance the equation and use diminutional analysis, but really your answer is gonna be like off by very little, also, because of rounding
aaronq
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did they give you the concentration of the acid?
eli123
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no
eli123
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all we had to do was place a piece of magnesium inside a buret that contained hcl and this was then converted and all the hcl migrated down to the bottom of the buret
eli123
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basically the magnesium was reacting with the hcl
eli123
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and after that we had to find the values in that table
aaronq
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hm well they really wouldn't need to tell you, do you know what a limiting reagent is?
eli123
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limiting reactant?
aaronq
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yep, it's the same thing.
you could use the number of moles of that and translate it to moles of H2 produced
2Mg (s) + 2HCl -> H2 + 2MgCl (aq)
0.001439 moles of Mg/2 = x moles of H2/1
moles of H2 = 0.0028794734677088
thats the theoretical yield
eli123
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well that makes more sense because thats what we are currently learning in class haha
aaronq
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oh wait sorry i multiplied, it should be
moles of H2 = 0.0007198683669272
aaronq
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haha oh true, well maybe let me know next time? haha
eli123
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sorry!!!
eli123
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so in other words, to find lets say trial #1
eli123
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where the volume of hydrgen gas produced was 37.00mL and the moles of h2 gas produced were 0.002 mol
aaronq
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you would convert the mass of Mg to moles
then using the balanced equation, build a relationship between H2 and Mg using their respective coefficients..
this just means:
moles of Mg/coefficient=moles of H2/coefficient
eli123
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oh my god
eli123
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moles of mg that reacted?
eli123
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coefficient is 2?
aaronq
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yeah because thats your limiting reactant..
aaronq
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for Mg is 2 for H2 is 1
aaronq
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wait no i balanced it wrong because it's
Mg + 2HCl -> MgCl2 + H2
aaronq
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so it's coefficient is 1 for both
aaronq
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sorry, i'm like lost today idk why
eli123
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i dont get it
aaronq
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Mg (s) + 2HCl -> H2 + MgCl2
0.001439 moles of Mg/1 = x moles of H2/1
moles of H2 = 0.001439
thats the theoretical yield
which is close to what you would get if you just divided the volume of gas produced by 22.414 L
eli123
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thats the ratio?
aaronq
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yep the ratio is 1:1
eli123
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but what value would i put?
aaronq
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i would go with the theoretical since thats what you're doing in class
eli123
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0.001439?
aaronq
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yeah, even round up to 0.0015
aaronq
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or down to 0.0014
eli123
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ok wait where did the 0.001439 come from>?
aaronq
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its the number of moles of magnesium
eli123
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ooooh yes thats my 0.001
aaronq
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haha yep, i mean round to whatever you think it's best, or whatever they're teaching you at school
eli123
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but for the ratio isnt moles of h2 divided by moles of magnesium? so 0.002h2/0.001mg
aaronq
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which ratio?
eli123
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ratio of volume displaced to moles h2 gas prodced
aaronq
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well the amount of moles of H2 produced is equal to the moles of Mg that reacted according to the stoichiometric ratios of the equation...
so to answer your question it would be
37 mL/0.001 moles.. which is what you had before, if i remember correctly
eli123
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37 being the volume of hydrogen gas produced and 0.001 moles of magnesium reacted?
aaronq
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well, technically yes, but the question is asking for the volume displaced divided by the moles of H2 produced..which are the same values
eli123
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oh so volume of liquid displaced divided by the moles of h2
eli123
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and thats the ratio?
eli123
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wow i feel stupid
aaronq
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that they're asking you for in the question yes.
They're basically seeing how close that value is to the ideal gas constant volume
of 1 mole =22.414 L, since
37mL/0.0014 mol = 26428.5714285714285714 mL/mol
26.428 L/mol
so it's close if you use more than 0.001, which i think you should do lol
eli123
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oh my i get the big number again hahaha
aaronq
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haha don't feel stupid
aaronq
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your friend in the other thread used 0.00144 moles
eli123
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0.0014 is in the correct significant figures? i keep using 0.001
aaronq
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hm idk though if you don't use that your error is pretty large
eli123
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exactly
eli123
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but look when i divide 0.035/ 24.31
aaronq
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0.0014397367338544
eli123
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and the right sig figures are four so you round it to 0.001?
aaronq
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zeros dont count though
aaronq
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well zeros on the left side
eli123
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oh..
eli123
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still..
eli123
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so now its 3 sig figures?
aaronq
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i would use 2 sig digits b/c 0.035 .. and have 0.0014
eli123
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so only 14 are significant
aaronq
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yes
eli123
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what about 0.016/24.31
eli123
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0.0066?