50.0 g of n2o4(g) is placed in a 2.0 l evacuated flask and the system is allowed to reach equilibrium according to the reaction: n2o4(g) ⇄ 2 no2(g) kc = 0.133 after the system reaches equilibrium, 5.00 g of no2(g) is injected into the vessel, and the system is allowed to equilibrate once again. calculate the mass of n2o4 in the final equilibrium mixture.
Answers: 3
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;O: 
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in that sample of 
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in that 
container: 
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be 
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.![\displaystyle \frac{[\rm NO_2]^{2}}{[\rm N_2O_4]} = \rm K_{c} = 0.133](/tpl/images/0308/9684/7a7f2.png)
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.![\rm [N_2O_4] \approx 0.271710 - 0.0798672 = \rm 0.191843 \;mol\cdot L^{-1}](/tpl/images/0308/9684/a5112.png)
.![\rm [NO_2] \approx 2 \times 0.0798672 = 0.159734\; mol\cdot L^{-1}](/tpl/images/0308/9684/f286e.png)
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of ![\rm[NO_2]](/tpl/images/0308/9684/e223a.png)
if it was added to an evacuated 
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will become 
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be 
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for if that would lead to a negative value for the concentration of 
.![[{\rm N_2O_4}] = 0.703154 + 2(-0.232758) = \rm 0.237638\; mol\cdot L^{-1}](/tpl/images/0308/9684/8370c.png)
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50.0 g of n2o4(g) is placed in a 2.0 l evacuated flask and the system is allowed to reach equilibriu...
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