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Stoichiometric Calculations

8. The mass o...

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8. The mass of N2F4 produced by the reaction of 1 mole of Nh3 and 2.5 mole of F2 is 35.6 g. Hence,the percentage yield of the given reaction is?

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Q. In the reaction,

2 N H_{3} (g) + 5 F_{2} \to N_{2} F_{4} + 6 H F

3.56 g N_{2} F_{4}

is obtained by mixing 2 g

N H_{3}

F_{2}

. The percentage yield of the production is:

Q. For the reaction,

2 F e (N O_{3})_{3} + 3 N a_{2} C O_{3} \to F e_{2} (C O_{3})_{3} + 6 N a N O_{3}

initially 2.5 mole of

F e (N O_{3})_{2}

and 3.6 mole of

N a_{2} C O_{3}

are taken. If 6.3 mole of

N a N O_{3}

is obtained then % yield of given reaction is:

9

D

14

E

are allowed to react in a closed vessel according to given reactions. Calculate the number of moles of

B

formed at the end of the reaction, if

4

G

are present in the reaction vessel. (Percentage yield of the reaction is mentioned in the reaction).
Step 1:

3 D + 4 E 80 % - - \to 5 C + A

3 C + 5 G 50 % - - \to 6 B + F

Q. For the reaction,

F e (N O_{3})_{3} + N a_{2} C O_{3} \to F e_{2} (C O_{3})_{3} + N a N O_{3}

Initially if 2.5 moles of

F e (N O_{3})_{2}

and 3.6 moles of

N a_{2} C O_{3}

is taken. If 6.3 moles of

N a N O_{3}

is obtained then

%

yield of given reaction will be:

(N_{2} H_{4})

can be produced according to the following reaction:

C l N H_{2} + 2 N H_{3} \to N_{2} H_{4} + N H_{4} C l

C l N H_{2}

is reacted with excess

N H_{3}

, find the actual number of molecules of

N_{2} H_{4}

produced if the percentage yield of the given reaction is 76%.

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