what would be the weight of slaked lime required to decompose 8 gram of ammonium chloride
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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 g
thus, 5 g of ammonium chloride will be decomposed by slaked lime = 74*5/107
= 3.45g
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