Question

In the reaction
`2NH_(3)(g)+5F_(2) to N_(2)F_(4)+6HF`
3.56g `N_(2)F_(4)` is obatained by mixing 2g `NH_(3) and 8g F_(2)`. The percentage yield of the production is:

Answer 1

Correct Answer - a
`underset(34g)(2NH_(3)(g))+underset(190g)(5F_(2)) to underset(104g)N_(2)F_(4)+6HF`
`"Amount of "N_(2)F_(4)"formed by 2g "NH_(3)=(104)/(34)xx2=6.12g`
`"Amount of "N_(2)F_(4)"formed by 8g "F_(2)=(104)/(190)xx8=4.38g`
`N_(2)F_(4)` will be limiting and actual amount of the product is 3.56g
`"% yield"=("Actual amount of product")/("Calculated amount of product")xx100`
`=(3.56)/(4.38)xx100=81.28%`