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Whatmass of slaked lime would be required to decompose completely 4 grams of ammonium chloride and what would be the mass of each product? |
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Answer» Solution :The equation representing the decomposition of `NH_(4)CL` by slaked time, i.e., `Ca(OH)_(2)` is : `{:(Ca(OH)_(2),+,2NH_(4)Cl,rarr,CaCl_(2),+,2NH_(3),+,2H_(2)O),(40+2(1+16),,2(14+4+35.5),,40+2xx35.5,,2(14+3xx1),,2(2xx1+16)),(=74g,,=107g,,=111g,,=34g,,=36g):}` (i) To calculate the mass of `Ca(OH)_(2)` required ot decompose 4 g `NH_(4)Cl`. From the above equation, 107 g of `NH_(4)Cl` are decomposed by 74 g of `Ca(OH)_(2)` THEREFORE` 4G` of `NH_(4)Cl` will be decomposed by `Ca(OH)_(2)=(74)/(107)xx4=2.766 g` Thus, the mass of slaked lime required = 2.766 g. (ii) To calculate the slaked lime required = 2.766 g. 107 g of `NH_(4)Cl` when reacted with `Ca(OH)_(2)` produce 111 g of `CaCl_(2)`. `therefore" 4 g of "NH_(4)Cl` when reacted with `Ca(OH)_(2)` produce `CaCl_(2)=(111)/(107)xx4=4.15 g` Hence, the mass of `CaCl_(2)` produced = 4.15 g. (iii) To calculate the mass of `NH_(3)` produced. 107 g of `NH_(4)Cl` react with `Ca(OH)_(2)` to give 34 g of `NH_(3)`. `therefore" 4g of "NH_(4)Cl` react with `Ca(OH)_(2)` to give produce `NH_(3)=(34)/(107)xx4=1.271 g` Hence, the mass of `NH_(3)` produced = 1.271 g. (iv) To calculate the mass of `H_(2)O` formed. 107 g of `NH_(4)Cl` react with `Ca(OH)_(2)` to yield 36 g of `H_(2)O` `therefore` 4 g of `NH_(2)Cl` react with `Ca(OH)_(2)` to yield `H_(2)O=(36)/(107)xx4=1.3458 g` So the mass of `H_(2)O` formed = 1.3458 g. |
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