Formation of polyethylene from calcium carbide takes place as follows: \[Ca{{C}_{2}}+2{{H}_{2}}O\to Ca{{(OH)}_{2}}+{{C}_{2}}{{H}_{2}}\]\[{{C}_{2}}{{H}_{2}}+{{H}_{2}}\to {{C}_{2}}{{H}_{2}}\] \[n{{C}_{2}}{{H}_{2}}\to {{(-C{{H}_{2}}-C{{H}_{2}}-)}_{n}}\] The amount of polyethylene obtained from 64.1 kg of\[Ca{{C}_{2}}\] is

Formation of polyethylene from calcium carbide takes place as follows:

1.	Formation of polyethylene from calcium carbide takes place as follows: \[Ca{{C}_{2}}+2{{H}_{2}}O\to Ca{{(OH)}_{2}}+{{C}_{2}}{{H}_{2}}\]\[{{C}_{2}}{{H}_{2}}+{{H}_{2}}\to {{C}_{2}}{{H}_{2}}\] \[n{{C}_{2}}{{H}_{2}}\to {{(-C{{H}_{2}}-C{{H}_{2}}-)}_{n}}\] The amount of polyethylene obtained from 64.1 kg of\[Ca{{C}_{2}}\] is
A.	7 kg
B.	4 kg
C.	21 kg
D.	28 kg
Answer» C. 21 kg

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