A 20 L flask contains 4.0 gm of O_(2) & 0.6 gm of H_(2) at 100^(@)C. If the contents are allowed to react to form water vapors at 100^(@)C,find the contents of flask and there partial pressures.

A 20 L flask contains 4.0 gm of O_(2) & 0.6 gm of H_(2) at 100^(@)C. I

1.	A 20 L flask contains 4.0 gm of O_(2) & 0.6 gm of H_(2) at 100^(@)C. If the contents are allowed to react to form water vapors at 100^(@)C,find the contents of flask and there partial pressures.
Answer» Solution : `H_(2)` reacts with `O_(2)` to form water `[H_(2)O_((g))]` `2H_(2(g))+O_(2(g)) rarr 2H_(2)o_((g))` `rArr 2` MOLES of `H_(2) = ` 1 mole of `O_(2)` = 2 moles of `H_(2)O` Here masses of `H_(2)` and `O_(2)` are given, so one of them can be in excess. So first check out which of the reactants is in excess. Now, Moles of `O_(2)= 4//32 = 0.125` and Moles of `H_(2) = 0.6 //2 = 0.3` SINCE 1 mole of `O_(2)= 2 `moles of `H_(2)` `rArr 0.125` moles of `O_(2) -= 2 xx 0.12 5 ` moles of `H_(2)` i.e., 0.25 moles of `H_(2)` are USED , so `O_(2)` reacts completelywhereas `H_(2)` is in excess. `rArr `Moles of `H_(2)` in excess `= 0.3 - 0.25 = 0.05` moles. Also, 2 moles of `H_(2) -= 2` moles of `H_(2)O` `rArr 0.25` moles of `H_(2) -= 0.25` moles of `H_(2)O` are produced. `rArr` Total moles after the reaction =0.05 ( moles of `H_(2) + 0.25` ( moles of `H_(2) O)=0.3` `rArr ` The total PRESSURE `P_("Total")` at the end of reaction is given by `:` `P_("Total")= (nRT)/( V) = (0.3 xx 0.0821 xx 373)/( 20) = 0.459 atm` Now partial pressure of A = mole FRACTION of `A xx P_("total")` `rArr P_(H_(2)) = 0.05 // 0.3 xx 0.459 = 0.076 ` atm `rArr P_(H_(2)O) = 0.25 // 0.3 xx 0.459 =0.383 ` atm