| \[2N{{H}_{3}}+5{{F}_{2}}\to {{N}_{2}}{{F}_{4}}+6HF\] |
A) 79.0
B) 71.2
C) 84.6
D) None of these
Correct Answer: D
Solution :
[d] \[2N{{H}_{3}}+5{{F}_{2}}\xrightarrow{{}}{{N}_{2}}{{F}_{^{4}}}+6HF\] \[2\times 17g\,N{{H}_{3}}\] gives 66 g \[{{N}_{2}}{{F}_{4}}\] 2 g will give\[-\frac{66}{34}\times 2=3.88g\,{{N}_{2}}{{F}_{4}}\] % yield \[=\frac{3.56}{3.88}\times 100=91.75%.\]
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