Question

For the following reaction, 6.99 grams of hydrogen gas are mixed with excess nitrogen gas. The reaction yields 27.7 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 ?

%

Answer 1

What is the Maximum amount of Agl, in grams, that can be produced in the reaction
Amount Nal reacting = 0.348
Theoetical Agl formed = 0.348

Answer 2

a student conduct an expirement by reacting 52.1 g Nal with an excess of AgNO3. The reaction yields 60.468 Agl. What is the percent yield.

AgNO3 (aq)+ Nal(aq)= AgI(s)+NaNO3(aq)

Answer 3

Can you help me aswer this

Answer 4

First moles of hydrogen =6.99/2

3 H2= 2 NH3 6.99/2= X MOLES NH3, which is 2.33 moles


moles of NH3=mass/MM of NH3

2.33=mass/17 theoritical mass 0f NH3 =2.33*17=39.61 grams

A)39.61,theoritical value as you can see from the equation above so called theory

B)percent yield=(actual /theoritical)*100

now actual =27.7 G
theoritical =39.61G from a baove


therefore P%=(27.7/39.61)*100=69.93183%

Answer 5

i am sorry this is the answer
69.9318%

Answer 6

please i dont think a percent yield can actually be more than a hundred

Answer 7

moles of NaI = 52.1/23+127 which is 325.625 E-3

MASS OF AgI = 325.625E-3 * (196+127) which is 105.176875 grams which is the theoritical value

the actual value is 60.468

therefore % yield= (60.468/105.176875 ) *100

which is 57.4917%

Answer 8

pleaase when i dont calculate well ,forgive me it is because i get tired sometimes from anatomy i am reading

Answer 9

OHH YOUR IN ANATOMY

mE TO

Answer 10

what school do you go to