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what would be the weight of the slaked lime requied IO decompose S.O g of ammonium chloride ?
In the reaction given below, one can see that 1 mole of slaked lime is required to completely decompose 2 moles of ammonium chloride.107 g ammonium chloride is decomposed be slaked lime = 74 gthus, 5 g of ammonium chloride will be decomposed by slaked lime = 74*5/107= 3.45g
Simply learn that the ratio of sake lime to amoniya calorie is 74 ÷ 107 (g) use this ratio to find the amount of amonium calorie and also sake lime
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