$1\mathrm{.}2$ Calcium carbide is considered to be a potential fuel in a basic oxygen

converter and would be expected to burn into CaO and CO or C$0?,$ depending

upon the conditions. If the heat required to raise steel scrap to $1600\backslash$deg C

$(1873$ K$)$ is $333$ kcal$/$kg $(1,393\mathrm{.}27$ hi$/$kg$)$ of scrap, calculate how many kg of

steel scrap can be charged per $1000$ kg of CaC$2$ when

$($ i $)$ all CaC$2$ is consumed to form CO$,$

$($ i i $)$ all CaC$2$ is consumed to form C$02,$ and

$($ i i i $)60\%$ of CaC$2$ is u t i l i z ed to produce C$02$ and rest for CO$.$

Assume that reaction takes place at $25\backslash$deg C $(298$ K$).$

Given: $($i$)\backslash$Delta $\backslash$Eta $\backslash$deg $298$ CaC $=$ ~ $14$ kcal$/$mole $(-58\mathrm{.}99$ kJ$/$mol$).$

$($ i i $)\backslash$Delta $\backslash$Eta $\backslash$deg $298$ Ca$0=-151\mathrm{.}8$ kcal$/$mole $(-635\mathrm{.}13$ kJ$/$mol$).$

$($ i i i $)\backslash$Delta $\backslash$Eta $\backslash$deg $298$ c o $=-26\mathrm{.}42$ kcal$/$mole $(-110\mathrm{.}54$ kJ$/$mol$).$

$($iv$)\backslash$Delta $\backslash$Eta $\backslash$deg $298$ CO $=-94\mathrm{.}05$ kcal$/$mole $(-393\mathrm{.}50$ kJ$/$mol$).$

$($v$)$ Atomic weights of Ca and C are $40$ and $12$ respectively.

The answers are:

$($i$)8940$ kg

$($ii$)15285$ kg

$($iii$)12745$ kg