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A 2.0 L container at `25^(@)C` contains 1.25 mol of oxygen and 3.3 mol of carbon. (a) What is the initial in the flask ? (b) If carbon and oxygen and oxygen react as completely as possible to form CO, what will be the final pressure in the container ? |
Answer» (a) V=2.0 L, T=298 K, n=1.25 mol (because only gas will exert pressure). Hence, PV=nRT or `P=(nRT)/(V)=((1.25" mol")(0.0821" L atm "K^(-1)mol^(-1))(298 K)/(2.0" L")=15.3" atm"` (b) The reaction will be :`C(s)+(1)/(2)O_(2)(g) to CO(g)` As 1 mol of C reacts with `(1)/(2)` mol of `O_(2)`, the limiting reactant will be 1.25 mol of `O_(2)`. Thus, `(1)/(2)` mol of `O_(2)` produces CO=1 mol. `:.1.25` mol of `O_(2)` will produce CO=2.50 mol. Now, n(gaseous)=2.50 mol. Hence, `P=(nRT)/(V)=((2.50" mol")(0.0821" L atm "K^(-1)mol^(-1)(298"K"))/(2.0" L ")=30.6" atm "`. |
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