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