In a reaction vessel, 0.600 mol of Ba(NO 3) 2(s) and 0.300 mol of H3 PO4 (aq) are combined with deionizedwater to a final volume of 2.00 L. The reaction represented below occurs.3 Ba(NO3) 2(aq) + 2 H3 PO4 (aq) Ba3 (PO4 )2 (s) + 6 HNO3 (aq)(i) Calculate the mass of Ba 3(PO4 )2 (s) formed.(ii) Calculate the pH of the resulting solution.(iii) What is the concentration, in mol L–1 , of the nitrate ion, NO3– (aq), after the reaction reachescompletion?
Question
In a reaction vessel, 0.600 mol of Ba(NO 3) 2(s) and 0.300 mol of H3 PO4 (aq) are combined with deionizedwater to a final volume of 2.00 L. The reaction represented below occurs.3 Ba(NO3) 2(aq) + 2 H3 PO4 (aq) Ba3 (PO4 )2 (s) + 6 HNO3 (aq)(i) Calculate the mass of Ba 3(PO4 )2 (s) formed.(ii) Calculate the pH of the resulting solution.(iii) What is the concentration, in mol L–1 , of the nitrate ion, NO3– (aq), after the reaction reachescompletion?
Solution
(i) To calculate the mass of Ba3(PO4)2 formed, we first need to determine the limiting reactant. The stoichiometry of the reaction tells us that 3 moles of Ba(NO3)2 react with 2 moles of H3PO4 to form 1 mole of Ba3(PO4)2.
We have 0.600 mol of Ba(NO3)2 and 0.300 mol of H3PO4. The ratio of Ba(NO3)2 to H3PO4 is 0.600/0.300 = 2. This is greater than the stoichiometric ratio of 3/2 = 1.5, so H3PO4 is the limiting reactant.
Since 2 moles of H3PO4 produce 1 mole of Ba3(PO4)2, 0.300 mol of H3PO4 will produce 0.300/2 = 0.150 mol of Ba3(PO4)2.
The molar mass of Ba3(PO4)2 is 601.93 g/mol, so the mass of Ba3(PO4)2
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