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A 2.0 L container at 25^(@)C contains 1.25 mol of oxygen and 3.3mol 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» <html><body><p></p>Solution :(a) V=2.0 L, T=298 <a href="https://interviewquestions.tuteehub.com/tag/k-527196" style="font-weight:bold;" target="_blank" title="Click to know more about K">K</a>, n=1.25 mol (because only <a href="https://interviewquestions.tuteehub.com/tag/gas-1003521" style="font-weight:bold;" target="_blank" title="Click to know more about GAS">GAS</a> will exert pressure). Hence, <br/> PV=nRTor`P=(<a href="https://interviewquestions.tuteehub.com/tag/nrt-1109890" style="font-weight:bold;" target="_blank" title="Click to know more about NRT">NRT</a>)/(V)=((1.25" mol")(0.0821" L atm "K^(-1)mol^(-1))(298 K)/(2.0" L")=15.3" atm"` <br/> (<a href="https://interviewquestions.tuteehub.com/tag/b-387190" style="font-weight:bold;" target="_blank" title="Click to know more about B">B</a>) The reaction will be :`C(s)+(1)/(2)O_(2)(g) to CO(g)` <br/> 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. <br/> `:.1.25` mol of `O_(2)` will produce CO=2.50 mol. <br/> Now, n(<a href="https://interviewquestions.tuteehub.com/tag/gaseous-467815" style="font-weight:bold;" target="_blank" title="Click to know more about GASEOUS">GASEOUS</a>)=2.50 mol. Hence, <br/> `P=(nRT)/(V)=((2.50" mol")(0.0821" L atm "K^(-1)mol^(-1)(298"K"))/(2.0" L ")=30.6" atm "`.</body></html>


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