A mixture of 0.10 mol of NO, 0.050 mol of H2. and 0.10 mol of H2O is placed in a

Question

A mixture of 0.10 mol of NO, 0.050 mol of H2. and 0.10 mol of H2O is placed in a 1.0-L vessel at 300 K. The following equilibrium is established 2NO(g) +2N2(9) + 2H20(9) At equilibrium NO 0.062M PartA Calculate the equilbrium concentration of H2 Express your answer using two significant figures 1.2x10-2 M Previous Answers Correct Part B Calculate the equilibrium concentration of N2 Express your answer using two significant figures. 1.9x102 M Previous Answers Correct Part C Calculate the equilbrium concentration of H2O 0.138 M Correct

Explanation / Answer

2NO(g) + 2H2(g) ------------> N2(g) + 2H2O(g)
I    0.1      0.05                          0.1
C   -2*0.019 -2*0.019              0.019   2*0.019
E    0.062     0.012                0.019   0.138
  
       [NO] = 0.062M = 6.2*10^-2M
       [H2] = 0.012M = 1.2*10^-2M
       [N2] = 0.019M = 1.9*10^-2M
       [H2O] = 0.138m = 1.38*10^-1M

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