A) \[C{{O}_{2}}=300mL;CO=400mL\]
B) \[C{{O}_{2}}=0.0mL;CO=700mL\]
C) \[C{{O}_{2}}=200mL;CO=500mL\]
D) \[C{{O}_{2}}=350mL;CO=350mL\]
Correct Answer: A
Solution :
\[C{{O}_{2}}+C\xrightarrow[{}]{{}}2CO\] Stoichoimetry ratio is 1 : 2 AT STP, P=1 atm, T=273 K, R = 0.0821 Initial moles of \[C{{O}_{2;}}n(C{{O}_{2}}\text{initial})=\frac{PV}{RT}\] \[=\frac{1\times 0.5}{0.0821\times 273}=0.022\,\text{mole}\] In final mixture no. of moles; \[n(C{{O}_{2}}/CO\,\text{mixture)}\] \[=\frac{1\times 0.7}{0.0821\times 273}=0.031\] Increase in volume is by = 0.031 - 0.022 = 0.009 mole of gas Final no. of moles of CO i.e. \[{{n}_{(CO\,final)}}\] \[{{n}_{(CO\,final)}}=2{{n}_{(C{{O}_{2}}\,initial)}}-{{n}_{(C{{O}_{2}}final)}}\] \[=2(0.022)-{{n}_{(C{{O}_{2}}final)}}\] ?(i) \[{{n}_{(CO\,final)}}=0.044-2{{n}_{(C{{O}_{2}}final)}}\] ?(ii) \[\therefore \]Now,\[{{n}_{(COfinal)}}+{{n}_{(COfinal)}}=0.031\] \[{{n}_{(C{{O}_{2}}final)}}=0.031-{{n}_{(CO\,final)}}\] Substituting (ii) in eq. (i) \[{{n}_{(CO\,final)}}=0.044-[0.031{{-}_{n(CO\,final)]}}\] \[{{n}_{(CO\,final)}}=0.044-0.062+2{{n}_{(CO\,final)}}\] \[{{n}_{(CO\,final)}}=0.018mol.\] Volume of \[CO=V=\frac{nRT}{P}=\frac{0.018\times 0.0821\times 273}{1}\] \[=0.40\]Litre and volume of \[C{{O}_{2}}=0.7\]litre ? 0.4 litre = 0.3 litre \[\therefore \]\[C{{O}_{2}}=300mL\,CO=400mL\]
You need to login to perform this action.
You will be redirected in
3 sec
