Question

a student conduct an expirement by reacting 71.3 g Nal with an excess of AgNO3. The reaction yields 75.940 Agl. What is the percent yield.

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

Answer 1

please apply the procedure with the 52.1 value here then you give me the answer to check

Answer 2

I think I can help!:)

Answer 3

i made acorrection to the first one with a) and b) so please check it out a bit

Answer 4

@jadehoff thank you

Answer 5

Your welcome, Well you have to do what @telijahmed said to do. :)

Answer 6

pply the procedure with the 52.1 value

Answer 7

i am on it

Answer 8

First step

calculate the theoritical mass of AgI and this is the procedure

calculate the moles of NaI=71.3/23+127 which is 445.623 *10^-3 moles

the ratio of AgI to NaI is 1:1 WHICH MEANS THAT THE MOLES OF AgI would also be 445.623 *10^-3 moles

calculate now the theoritical mass of AgI and it is given by the formula

MASS=MOLES*MOLAR MASS
MASS OF AgI = 445.623 *10^-3 *(127+197)
the value is 144.3825 grams

now that you have

THEORITICAL MASS = 144.3825

ACTUAL MASS=75.940

PERCENT YIELD=(ACTUAL M/THEORITICAL MASS) * 100

=(75.940/144.3825)*100=52.5964019%

Answer 9

52.5964019%

Answer 10

double check if i fit in the right values and if so then this is the right answer

Answer 11

please get back to me to questions about the concept because i am about to go on a long break from the forum

Answer 12

this is great

Answer 13

thanks