For the following reaction, 6.33 grams of hydrogen gas are mixed with excess nitrogen gas. The reaction yields 25.7 grams of ammonia. N2 (g) + 3 H2 (g) = 2 NH3 (g) (1) What is the theoretical yield of ammonia ? grams (2) What is the percent yield for this reaction ? %
so for theoretical yield you take a look at what you put in and how much you get out. N2 is14g/mol. ammonium is (14+3)*2 (number of molecules we get after the reaction).
for actual yield you have to multiply 2 molecules of ammonia by 25,7 gramm.
for percentage yield: actual yield/theoretical yield*100
the actual yield is what they gave you 25.7 grams The theoretical is what you find through the use of the stoichiometric coefficients
so use this percent yield = (actual yield/theoretical yield)*100
lol yep
For the following reaction, 6.11 grams of oxygen gas are mixed with excess nitrogen monoxide. The reaction yields 14.1 grams of nitrogen dioxide. 2 NO (g) + O2 (g) = 2 NO2 (g) (1) What is the theoretical yield of nitrogen dioxide ? grams (2) What is the percent yield for this reaction ?
the same way you did in the first question
ok, convert the mass of oxygen to moles
nope, use: moles=mass/molar mass
okay this is ver diffucult for me so please bear with me
can you please help me and THEN i WILL KNOW HOW TO DO IT
what are you having difficulty with?
the moles=mass/molar mass
it's an algebraic expression, you plug in your values and then solve for your unknown. just like this: x=y/z whats the molar mass of oxygen gas? (remember that it's 2 atoms of oxygen per molecule)
The molar mass of oxygen gas is 32.
sorry, yes thats right, now plug in what you have into the equation and solve for moles.
so 32 is the therotical yield
no, you first have to find the moles of oxygen you have
moles = mass/molar mass
what did you do?
did you find my solution helpful last time
yes I did find it helpfull
nope, moles = mass / molar mass whats the value for mass, whats the value to molar mass
molar mass is 32
and the mass?
how do i find that
it's given in the question
no were dealing with oxygen, 6.11 grams now, plug it into moles = mass / molar mass
moles=61.1/32
close, moles = 6.11/32 what do you get?
.1909375
now you're goin to use the balanced equation 2 NO (g) + O2 (g) = 2 NO2 (g) \[\frac{ molesO _{2} }{ 1 }=\frac{ moles of NO _{2} }{ 2 }\] plug into this and solve for for moles of NO2
notice that you i divided each species by it's coefficient in the reaction
.381875
is that the anser
that is what a I got
now you have to convert back to grams of NO2
look at what i have attached very clearly please ,you just caused me some ink
yes that is the answer for the theoritical value of NH3
the answer to the 1st question
now i ll do the second part and scan
oh is see
use the same procedure
he's right, you use the same procedure for all of these problems. you got 38 grams of NO2?
how did you get 38?
how did you get 38 grams of NO2?
I dont know what I di
lol okay, well thats not good. you had how many moles of NO2?
.381875
moles of oxygen = 6.11/32 if 2 NO2 = 1 O2 X NO2 = 6.11/32 MOLE OF NO2= MASS/MM MASS OF NO2=(6.11/32)*2 *46 17.56625 GRAMS
ok that is the answer
yep, but honestly telijahmend you're not doing her a favour by just typing out the answer
thanks you can you help with one more so I UNDERTAND IT A LITTLE
percent yield=actual yield/theoritical above now actual is 14.1 theoritical= above so go on from here @aaronq is right so do the algebra now
because this is very vey confusing to me
i got the anserw
telijamed
you have the formula so go ahead and just put in the values
then multiply by a 100 since it is percent yield
ok i am going to start over in adiifent page with a new question
I go the anser
is is 80.1 percent
(14.1/17.56625) *100 just go ahead and get the answer now
80
80.26 %
can you please help me on another questio
just go ahead and keep your periodic table by your side and it ll be fine
you ve got to put in so much effort
i know
but can you helpnother question like this
so I get understanding of this
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