1.

Formation of polyethylene from calcium carbide takes place as follows CaC_(2)+2 H_(2)O rarr Ca(OH)_(2)+C_(2)H_(2) C_(2)H_(2)+H_(2) rarr C_(2)H_(2) N(C_(2)H_(4)) rarr (-CH_(2)-CH_(2)-)_(n) The amount of polyethylene obtained from 64.1 kgCaC_(2)is

Answer»

7 kg
14 kg
21 kg
28 kg

Solution :MOLES of ` CaC_(2)=(64xx10(3))/(64 )~=1 xx 10^(3)`
`:.` From the balanced chemical equations, moles of `C_(2)H_(2)=` moles of `C_(2)H_(4)` =moles of `CaC_(2)=1 xx 10^(3)`
`:.` moles of polythene `=(1)/(n)xx 1 xx 10^(3)`
`:.`weightof polythene `=(1)/(n) xx 1 xx 28n` kg=28kg


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