reyrey389 - 22-2-2012 at 19:21
2NO + 2H2 --> N2 + 2H2O
The concentration of NO is 0.006 M and the concentration for H2 is 0.002M. How can I find the concentration of NO remaining when one-half of the
original amount of H2 has been consumed?
if you can also show me how it would be done if the reaction was not 1:1, that would help alot. thanks
ThatchemistKid - 22-2-2012 at 19:27
well, if 1/2 of the H2 that is there is consumed mole wise what happens to the NO via the reaction equation?
Hohenheim - 22-2-2012 at 19:38
If half of the H2 is consumed, that means one half of the total moles of H2 has been converted; given that 2 molecules of H2 is consumed with 2
molecules of NO, we can assume that the number of moles of H2, 0.002/2=.001 (half), that is converted is the same for NO. So, 0.006-0.001=0.005. That
is the concentration of NO.
If it is not 1:1, figure out how much of one reactant is used per molecule of the other in conversion. So if 3 molecules of NO are used per 2
molecules of H2, take 3/2* the moles of H2 consumed and subtract as shown earlier.
reyrey389 - 22-2-2012 at 19:54
Hohenheim: ty for replying. it makes more sense now, the reason i was confused is because for a somewhat similar problem, "A + 2B --> C
After 5.0 * 10^(-3) s has elapsed the concentration of A is 7.5 * 10^(-4). What is the concentration of C now" they did 1 - concentration of A.
i.e. 2.5 *10^(-4). However for the problem i posted here, we didn't subtract from 1.
Also for that question i put in quote would the concentration of B be 2*(7.5) or 2*(2.5)
Thanks!