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 :
[a] \[C{{O}_{2}}+C\xrightarrow{{}}2CO\] Stoichoimetry ratio is 1 : 2 AT STP, P = 1 atm, T= 273 K, R = 0.0821 Initial moles of \[C{{O}_{2}}.n(C{{O}_{2}}initial)=\frac{PV}{RT}\] \[=\frac{1\times 0.5}{0.0821\times 273}=~0.022\text{ }mole\] In final mixture no. of moles; \[n(C{{O}_{2}}/CO\]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) \[\therefore \,Now,\,{{n}_{(CO\,final)}}+{{n}_{(CO\,final)}}=0.031\] \[\,{{n}_{(C{{O}_{2}}\,final)}}=0.031-{{n}_{(CO\,final)}}\] ...(iii) Substituting (ii) in eq. (i) \[{{n}_{(CO\,final)}}=0.044-2[0.031]-{{n}_{(CO\,final)}}\] \[{{n}_{(CO\,final)}}=0.044-0.062+2{{n}_{(CO\,final)}}\] \[{{n}_{(CO\,final)}}=0.018\,mol.\] 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 \[\therefore C{{O}_{2}}=300mL,\,CO=400mL\]
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