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A sample of Mg was burnt in air to give a mixture of MgO and $Mg_3N_2$.The ash was dissolved in 60meq. of HCl and the resulting solution was back titrated with NaOH.12meq of NaOH were required to reach the end point.An excess of NaOH was added and the solution distilled.The ammonia released was then trapped in 10Meq of second acid solution .Back titration of this solution required 6Meq of the base.Calculate the percentage of Mg burnt to the nitride.

$(a)\;27.27\%\qquad(b)\;36\%\qquad(c)\;42\%\qquad(d)\;41.2\%$

1 Answer

Step 1:
$2Mg+O_2\rightarrow 2MgO$
Before reaction:
$2Mg\Rightarrow a$
$2MgO\Rightarrow 0$
After reaction:
$2Mg\Rightarrow 0$
$2MgO\Rightarrow a$
$3Mg+N_2\rightarrow Mg_3N_2$
Before reaction:
$3Mg\Rightarrow b$
$Mg_3N_2\Rightarrow 0$
After reaction:
$3Mg\Rightarrow 0$
$Mg_3N_2\Rightarrow \large\frac{b}{3}$
Step 2:
Now $(a+\large\frac{b}{3})$ millimole of MgO and $Mg_3N_2$ are present in the mixture.
$MgO+2HCl\rightarrow MgCl_2+H_2O$
$Mg_3N_2+8HCl\rightarrow 3Mgcl_2+2NH_4Cl$
Or the solution contains a millimole of $MgCl_2$ from MgO
b millimole & $\large\frac{2b}{3}$ millimole of $NH_4Cl$ from $Mg_3N_2$
$2a+\large\frac{8b}{3}$ millimole of HCl used .
2a for Mgo
$\large\frac{8b}{3}$ for $Mg_3N_2$
Meq of HCl=60-12=48
$2a+\large\frac{8b}{3}$$=48$
Step 3:
millimole of $NH_4Cl$=millimole of $NH_3$ liberated =millimole of HCl used for absorbing $NH_3$
$\large\frac{2b}{3}$$=4$ or $b=6$
$2a+\large\frac{8\times 6}{3}$$=48$
$a=16$
$\%$ of Mg used for $Mg_3N_2=\large\frac{6}{(6+16)}$$\times 100$
$\Rightarrow 27.27\%$
Hence (a) is the correct answer.
answered Nov 6, 2013 by sreemathi.v
 

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