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

9

Explanation

`{:(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`