2 × 0.20 M = 0.40 M = 0.40 mol/L.In 2.5 L of the solution, there are 2.5 L × 0.40 mol/L = 1.00 mol K+ ions.57 . B
In the balanced equation,
2 Na 3 PO 4 (aq) + 3 Mg(NO 3 ) 2 (aq) → Mg 3 (PO 4 ) 2 (s) + 6 NaNO 3 (aq).
Complete reaction requires 3 moles of Mg(NO 3 ) 2 for every 2 moles of Na 3 PO 4 , so in this
experiment, magnesium phosphate is the limiting reactant. Therefore, there will be some
excess phosphate ions left in solution after the reaction (option A is false). Option D looks
tempting, but when we added together equal volumes of the two solutions, the nitrate ion
concentration of the original magnesium nitrate solution was cut in half.58 . D
In the original sodium phosphate solution, the concentration of sodium ions is (3 × 0.10 M =
0.30 M). When equal volumes of the two solutions are combined, the concentration of sodium
ions is halved to 0.15 M. (Note that, in fact, you didn’t even need to balance the equation in
this problem to get the correct answer. It’s a good idea to read the question before balancing
the equation—but in most cases balancing will be required.)
59 . D
Reversing equation I (changing the sign of its ∆H value!) and adding it to equation II gives SO 2
as a reactant and SO 3 as a product, as in the combustion equation for SO 2 :
The combustion equation as given requires us to divide our equation (I), and its ∆H value by 4,
so we’re looking for an answer in the neighborhood of −200 kJ/mol (or do the math and get
−197.8 kJ).