Question

Calcium carbide is made in an electric furnace by the reaction:
$CaO+3C⟶Ca{C}_2+CO$
The crude product is usually $85\%$ $Ca{C}_2$ and $15\%$ unreacted $CaO$.
(a)The amount of $CaO$ with which we should start to produce $450$ kg of crude product is $x$ kg.
(b) Amount of $Ca{C}_2$ which this crude product will contain is $y$ kg.
The value of $x+y$ in a nearest integer is :

Answer 1

$CaO+3C⟶Ca{C}_2+CO$

molecular wt of $CaO$ = $56g/mol$

and $Ca{C}_2$= $64g/mol$

Product of the reaction has $85$% $Ca{C}_2$ and $15$% $CaO$

(i) a mol of $CaO$ will produce $0.85\times amolCa{C}_2$ and $0.15\times amolCaO$ in the product

Therefore in $450$ kg ($=450000g$) product we have:

$0.85\times a\times 64+0.15\times a\times 56=450000$

$a=7165.6$

Amount of $CaO$ used is $x=a\times 56/1000=401.27kg$

(ii) Amount of $Ca{C}_2$ formed is:

$y=0.85$ $\times 7165.6\times 64/1000=389.81kg$

$x+y=791kg$

Answer is $791kg$