hey u guys Plz help
Okay, let's write out the balanced equation first. \[1. H_{2(g)}+ I_{2(g)} \rightarrow 2HI_{g}\]
yes i got that
how do we find the limiting reagent?
Good question: it doesn't say how many moles/grams of each starting material we have
do you have this information?
wait there is a drawing on the bottom of the page, leeme attach it gimme a min
thats all the info i have
so we've got 7 molecules of H2 and 4 molecules of I2
yes
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.
how is it one to one ratio?
sorry im trying to make sure i understand this
one mole of hydrogen gas reacts with one mole of iodine to for 2 moles of HI
that's from our balanced chemical reaction
yes i get that but dont u look on the other side of the equation for your ratio as well?
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
oh ok i understand
so what do we do next
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. \[4~molecules~I_{2}*(\frac{ 2HI }{ I_{2} }) = 8~molecules~of~HI\]
for C @popcornfun we know that we have 3 molecules of HI remaining
and we must add 3 more molecules of I2 for the reaction to go to completion
@popcornfun
ok so for D do we use the info from C?
@Photon336
yes so you know you have 3 molecules of H2 left over
so you would need 3 more molecules of I2 to make the reaction go to completion.
ohhhh ok
Thank you so much :) @Photon336
No problem man anytime
i might bother u again in the future dude im apologizing in advance XD @Photon336
yeah don't worry ask away
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