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If `340 g` of a mixture of `N_(2)` and `H_(2)` in the correct ratio gas a `20%` yield of `NH_(3)`. The mass produced would be:A. `16 g`B. `17 g`C. `20 g`D. `68 g` |
Answer» `20%` yield of `NH_(3)` and thus `20%` of `340 g` `=(20xx340)/(100)=68 g` |
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