Question

for the reaction 2fe(no3)2 +3na2co3=fe2(co3)3+6nano3 initially 2.5 moles of fe(no3)2 and 3.6 mole of na2co3 is taken 6.3 mole of nano3 is obtained then find the yield %

Answer 1

2Fe(NO3)2 +3Na2CO3 = Fe2(CO3)3+6NaNO3 


From the equation, 2 moles of Fe(NO3)2  will require 3 moles of 3Na2CO3


∴ 2.5 moles of Fe(NO3)2 will require = (2.5 x 3)/2

                                                           = 3.75 moles of Na2CO3



But we have only 3.6 moles of Na2CO3, so Na2CO3 is the limiting reactant.


From the equation, 3 moles of Na2CO3 produce 6 moles of NaNo3


∴ 3.6 moles of Na2CO3 will produce = (3.6 x 6)/3


                                                           = 7.2 moles of NaNO3



Only 6.3 moles of NaNO3 has been produced


Percentage yield = (6.3/7.2) x 100


                            = 87.5%

Answer 2

Explanation:

2Fe(NO3)2 +3Na2cO3

Fe2(CO3)3+6NaNo3

From the equation, 2 moles of Fe(NO3)2

will require 3 moles of 3Na2co3

2.5 moles of Fe(NO3)2 will require

(2.5 x 3)/2

=3.75

moles of Na2CO3

But we have only 3.6 moles of Na2C03,

so Na2Co3 is the limiting reactant.

From the equation, 3 moles of Na2C03

produce 6 moles of NaNo3

3.6 moles of Na2C03 will produce

(3.6 x 6)/3

7.2

moles of NaNO3

Only 6.3 moles of NaNO3 has been

produced

Percentage yield = (6.3/7.2) x 100

= 87.5%