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What mass of slaked lime would be requir...

What mass of slaked lime would be required to decompose completely 4.28 g of ammonium chloride ?

Text Solution

Verified by Experts

The equation representing the decomposition of `NH_(4)Cl` by slaked lime, `Ca(OH)_(2)` is

From the above equation,
107 g of `NH_(4)Cl` are decomposed by= 74 g of `Ca(OH)_(2)`
`therefore` 4.28 g of `NH_(4)Cl` will be decomposed by
`=74/107xx4.28=2.96g` of `Ca(OH)_(2)`
Thus, the required mass of slaked lime is 2.96 g.
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