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28 g of `N_(2)` and 6 g of `H_(2)` were mixed. At equilibrium 17 g `NH_(3)` was produced. The weight of `N_(2)` and `H_(2)` at equilibrium are respectivelyA. `11 g`, zeroB. `1g`, `3`C. `14g`, `3g`D. `11g`, `3g` |
Answer» `{:(,,N_(2),+,3H_(2),hArr,2NH_(3)),(,,28/28=1,,6/2=3,,0"mole before reaction"),(,,1-1/2,,3-3/2,,17/17=1"mole after reaction"):}` `:. "Mole of" N_(2)=(1)/(2)` `Wt. of N_(2)=14 g` Mole of `H_(2)=(3)/(2)` Wt. of `H_(2)=(3)/(2)xx2=3g` |
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