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Chemistry 16 Online
OpenStudy (toxicsugar22):

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 ? %

OpenStudy (anonymous):

http://en.wikipedia.org/wiki/Yield_%28chemistry%29

OpenStudy (anonymous):

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).

OpenStudy (anonymous):

for actual yield you have to multiply 2 molecules of ammonia by 25,7 gramm.

OpenStudy (anonymous):

for percentage yield: actual yield/theoretical yield*100

OpenStudy (aaronq):

the actual yield is what they gave you 25.7 grams The theoretical is what you find through the use of the stoichiometric coefficients

OpenStudy (aaronq):

so use this percent yield = (actual yield/theoretical yield)*100

OpenStudy (aaronq):

lol yep

OpenStudy (toxicsugar22):

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 ?

OpenStudy (aaronq):

the same way you did in the first question

OpenStudy (aaronq):

ok, convert the mass of oxygen to moles

OpenStudy (aaronq):

nope, use: moles=mass/molar mass

OpenStudy (toxicsugar22):

okay this is ver diffucult for me so please bear with me

OpenStudy (toxicsugar22):

can you please help me and THEN i WILL KNOW HOW TO DO IT

OpenStudy (aaronq):

what are you having difficulty with?

OpenStudy (toxicsugar22):

the moles=mass/molar mass

OpenStudy (aaronq):

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)

OpenStudy (toxicsugar22):

The molar mass of oxygen gas is 32.

OpenStudy (aaronq):

sorry, yes thats right, now plug in what you have into the equation and solve for moles.

OpenStudy (toxicsugar22):

so 32 is the therotical yield

OpenStudy (aaronq):

no, you first have to find the moles of oxygen you have

OpenStudy (aaronq):

moles = mass/molar mass

OpenStudy (aaronq):

what did you do?

OpenStudy (anonymous):

did you find my solution helpful last time

OpenStudy (toxicsugar22):

yes I did find it helpfull

OpenStudy (aaronq):

nope, moles = mass / molar mass whats the value for mass, whats the value to molar mass

OpenStudy (toxicsugar22):

molar mass is 32

OpenStudy (aaronq):

and the mass?

OpenStudy (toxicsugar22):

how do i find that

OpenStudy (aaronq):

it's given in the question

OpenStudy (aaronq):

no were dealing with oxygen, 6.11 grams now, plug it into moles = mass / molar mass

OpenStudy (toxicsugar22):

moles=61.1/32

OpenStudy (aaronq):

close, moles = 6.11/32 what do you get?

OpenStudy (toxicsugar22):

.1909375

OpenStudy (aaronq):

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

OpenStudy (aaronq):

notice that you i divided each species by it's coefficient in the reaction

OpenStudy (toxicsugar22):

.381875

OpenStudy (toxicsugar22):

is that the anser

OpenStudy (toxicsugar22):

that is what a I got

OpenStudy (aaronq):

now you have to convert back to grams of NO2

OpenStudy (anonymous):

OpenStudy (anonymous):

look at what i have attached very clearly please ,you just caused me some ink

OpenStudy (anonymous):

yes that is the answer for the theoritical value of NH3

OpenStudy (anonymous):

the answer to the 1st question

OpenStudy (anonymous):

now i ll do the second part and scan

OpenStudy (anonymous):

oh is see

OpenStudy (anonymous):

use the same procedure

OpenStudy (aaronq):

he's right, you use the same procedure for all of these problems. you got 38 grams of NO2?

OpenStudy (aaronq):

how did you get 38?

OpenStudy (aaronq):

how did you get 38 grams of NO2?

OpenStudy (toxicsugar22):

I dont know what I di

OpenStudy (aaronq):

lol okay, well thats not good. you had how many moles of NO2?

OpenStudy (toxicsugar22):

.381875

OpenStudy (anonymous):

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

OpenStudy (toxicsugar22):

ok that is the answer

OpenStudy (aaronq):

yep, but honestly telijahmend you're not doing her a favour by just typing out the answer

OpenStudy (toxicsugar22):

thanks you can you help with one more so I UNDERTAND IT A LITTLE

OpenStudy (anonymous):

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

OpenStudy (toxicsugar22):

because this is very vey confusing to me

OpenStudy (toxicsugar22):

i got the anserw

OpenStudy (toxicsugar22):

telijamed

OpenStudy (anonymous):

you have the formula so go ahead and just put in the values

OpenStudy (anonymous):

then multiply by a 100 since it is percent yield

OpenStudy (toxicsugar22):

ok i am going to start over in adiifent page with a new question

OpenStudy (toxicsugar22):

I go the anser

OpenStudy (toxicsugar22):

is is 80.1 percent

OpenStudy (anonymous):

(14.1/17.56625) *100 just go ahead and get the answer now

OpenStudy (toxicsugar22):

80

OpenStudy (anonymous):

80.26 %

OpenStudy (toxicsugar22):

can you please help me on another questio

OpenStudy (anonymous):

just go ahead and keep your periodic table by your side and it ll be fine

OpenStudy (anonymous):

you ve got to put in so much effort

OpenStudy (toxicsugar22):

i know

OpenStudy (toxicsugar22):

but can you helpnother question like this

OpenStudy (toxicsugar22):

so I get understanding of this

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