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Chemistry 45 Online
OpenStudy (anonymous):

hey u guys Plz help

OpenStudy (anonymous):

OpenStudy (photon336):

Okay, let's write out the balanced equation first. \[1. H_{2(g)}+ I_{2(g)} \rightarrow 2HI_{g}\]

OpenStudy (anonymous):

yes i got that

OpenStudy (anonymous):

how do we find the limiting reagent?

OpenStudy (photon336):

Good question: it doesn't say how many moles/grams of each starting material we have

OpenStudy (photon336):

do you have this information?

OpenStudy (anonymous):

wait there is a drawing on the bottom of the page, leeme attach it gimme a min

OpenStudy (anonymous):

OpenStudy (anonymous):

thats all the info i have

OpenStudy (photon336):

so we've got 7 molecules of H2 and 4 molecules of I2

OpenStudy (anonymous):

yes

OpenStudy (photon336):

so finding the limiting reagent is easy: since this is a 1:1 molar ratio we would need to have an equal amount 7 molecules of hydrogen 7 molecules of I2 but we have 7 molecules of hydrogen gas 4 molecules of I2 so we know right away that I2 is limiting because that's going to run out first.

OpenStudy (anonymous):

how is it one to one ratio?

OpenStudy (anonymous):

sorry im trying to make sure i understand this

OpenStudy (photon336):

one mole of hydrogen gas reacts with one mole of iodine to for 2 moles of HI

OpenStudy (photon336):

that's from our balanced chemical reaction

OpenStudy (anonymous):

yes i get that but dont u look on the other side of the equation for your ratio as well?

OpenStudy (photon336):

no, the reason is that we want to figure out how much of each reactant we need. whatever reactant runs out first that's called the limiting reagent we then use that to figure out how much product we have

OpenStudy (anonymous):

oh ok i understand

OpenStudy (anonymous):

so what do we do next

OpenStudy (photon336):

we need seven molecules of iodine and seven molecules of hydrogen for the reaction to go to completion. but we only have four molecules of I2 so to figure out how much product we would get we would do this we would multiply the limiting reagent in this case iodine by the molar ratio to get our product. \[4~molecules~I_{2}*(\frac{ 2HI }{ I_{2} }) = 8~molecules~of~HI\]

OpenStudy (photon336):

for C @popcornfun we know that we have 3 molecules of HI remaining

OpenStudy (photon336):

and we must add 3 more molecules of I2 for the reaction to go to completion

OpenStudy (photon336):

@popcornfun

OpenStudy (anonymous):

ok so for D do we use the info from C?

OpenStudy (anonymous):

@Photon336

OpenStudy (photon336):

yes so you know you have 3 molecules of H2 left over

OpenStudy (photon336):

so you would need 3 more molecules of I2 to make the reaction go to completion.

OpenStudy (anonymous):

ohhhh ok

OpenStudy (anonymous):

Thank you so much :) @Photon336

OpenStudy (photon336):

No problem man anytime

OpenStudy (anonymous):

i might bother u again in the future dude im apologizing in advance XD @Photon336

OpenStudy (photon336):

yeah don't worry ask away

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