(b) 2.499 gms of coal sample was taken in silica crucible and heated in oven maintained at 110°C for one hour. The weight after heating was 2.368 gms. The same sample was analyzed for volatile matter and weight obtained was 1.75g, the sample was further treated to get fixed weight of 0.95 gms. Calculate the percentage of moisture, volatile matter, ash and fixed carbon for this sample

Given,

Weight of coal sample = 2.499 gms = $ w_1$

Weight of coal after heating at 110°C = 2.368 gms = $w_2$

Weight of coal after heating for Volatile Matter (V.M.) = 1.75 gms = $w_3$

Fixed weight obtained = 0.95 gms = $w_4 $

$ \implies \%\ moisture\ =(\frac{Loss\ in\ Weight}{Weight\ of\ coal } )\times100$

$ \implies {w_1-w_2 \over w_1}\times100$

$\implies {0.131 \over 2.499}\times100 =\ 5.24 \%$

$And,$

$\begin{aligned} \%\ V.M.\ = {Loss\ in\ weight\ due\ to\ V.M. \over Weight\ of\ coal} \times 100\end{aligned}$

$\begin{aligned} \ \ =\ {w_2-w_3 \over w_1}\times 100 \end{aligned}$

$\begin{aligned} \ \ =\ {2.368-1.75\over2.499}\times 100 =\ 24.7298 \% \end{aligned}$

$\begin{aligned} \%Ash\ = {Constant\ weight\ \over Weight\ of\ coal} \times 100 \end{aligned}$

$ \begin{aligned} \ \ =\ {w_4 \over w_1}\times 100 \end{aligned}$

$\begin{aligned} \ \ =\ {0.95\over2.499}\times 100 =\ 38.0152 \% \end{aligned}$

$ \% Fixed\ Carbon\ = 100\ -\ [\%moisture\ +\ \%V.M.\ + \%Ash]$

$\ \ =\ 100\ -\ [5.24\ + 24.72\ + 38.01]$

$\ \ =\ 100\ -\ [67.97]\ =\ 32.03 \%$

$ \therefore\ According\ to\ the\ question,\ we\ have\ $

$\% moisture\ =\ 5.24\% \\\\ \% Volatile\ Matter\ =\ 24.72\% \\\\ \% Ash\ =\ 38.01\% \\\\ \% Fixed\ carbon\ =\ 32.03\%\ \ \ Ans.$