1.

What is the maximum mass of H_(2)O (in gm) which can be obtained if total 42 gm of propyne & oxygen are subjected to combusion ?

Answer»


Solution :`{:(C_(3)H_(4(G)),+,4O_(2(g)),to,3CO_(2(g))+2H_(2)O_((l))),((40 GM)/(1/4"mole"),,128 gm,,1/2 "mole" =9 gm):}`
168 gm mixture should have 40 gm psopyne
42 gm mixture should have `40/168 xx42=10 gm`


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