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The weight of slaked lime required to decompose 8.0g $NH_4Cl$ is

$(a)\;5.53g\qquad(b)\;2.12g\qquad(c)\;15.52g\qquad(d)\;7.62g$

1 Answer

$2NH_4Cl+Ca(OH)_2\rightarrow CaCl_2+2NH_3+2H_2O$
$2\times$ 53.5g $NH_4Cl$=74g $Ca(OH)_2$
$\therefore 8gNH_4Cl=\large\frac{8\times 74}{2\times 53.5}$
$\Rightarrow 5.53g$
Hence (a) is the correct answer.
answered Jan 8, 2014 by sreemathi.v
 

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