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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» 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%` |
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