I believe that the best answer among the choices provided by the question is
<span>D.[HI] in experiment A equals 1/2[I2] in experiment B.</span>
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The ionic eqn is as follow:
1 Al(OH)3(s) + 3 H+(aq) + 3 NO3(-1) --> 1 Al(3+)(aq) + 3 NO3(-)(aq) + 3 H2O(l)
3moles of No3- ion on both sides cancels out to give the net ionic eqn:
1 Al(OH)3(s) + 3 H+(aq) --> 1 Al(3+)(aq) + 3 H2O(l)