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25. A mixture of 0.100 mol of NO, 0.0500 mol of H2, and 0.100 mol of H2O is placed in a 1.00-L vessel. The following equilibr
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Answer #1

ICE Table:

[NO] [H2] [H201 initial 0.1 0.05 0.1 change -2x -2x +2x equilibrium 0.1-2x 0.05-2x 0.1+2x

Given at equilibrium,

[NO] = 0.062

0.1-2x = 0.062

x = 0.0190

At equilibrium:

[H2] = 0.05-2x = 0.05-2*0.019 = 0.012 M

[H2O] = 0.1+2x = 0.1+2*0.019 = 0.138 M

[N2] = +1x = +1*0.019 = 0.019 M

Answer:

[H2] = 0.0120 M

[H2O] = 0.138 M

[N2] = 0.0190 M

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