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Ammonia gas can be prepared by following reaction : `CaO (s) + 2NH_(4) Cl (s) rarr 2NH_(3) (g) + H_(2)O (g) + CaCl_(2) (s)` If 112 gm of CaO and 224 gm of `NH_(4)Cl` are mixed and 17 gm of `NH_(3)` is formed thenA. Limiting reagent is `NH_(4)Cl`B. % yield of reaction is 25%C. Mass of steam formed is 18 gmD. Mass of `CaCl_(2)` formed is 55.5 gm |
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Answer» Correct Answer - B::D `{:(CaO (s),+,2NH_(4) Cl (s),rarr ,2NH_(3) (g),+,H_(2)O (g),+,CaCl_(2) (s),),(((112)/(56) gm),,((224)/(53.5) gm),,,,,,,),(= 2 "mol",,= 4.22 mol,,,,,,,):}` Moles of `NH_(3)` formed = 1 mol % yield `= (1)/(4) xx 100 = 25%` Moles of `H_(2)O` obtained `= xx (25)/(100) = (1)/(2) mol = 9 gm` Mass of `CaCl_(2) = 2 xx (25)/(100) = 111 = 55.5 gm` |
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