in a vessel, 0.6 mol Ba(NO3)2 and .3 mol H3PO4 are combined w deionized water to a final vol of 2 L.

3Ba(NO3)2 + 2H3PO4 --> Ba3(PO4)2 + 6HNO3

1. Calclate mass of Ba(PO4)2 formed

2. calcualate pH of HN03

3. What the concentation in mol/L of the nitrate ion after the rxn reaches complettion?

1. Calculate moles Ba3(PO4)2 formed from each reactant:

a. Using the coefficients in the balanced equation, convert moles Ba(NO3)2 to moles Ba3(PO4)2.
b. Same procedure, convert moles H3PO4 to moles Ba3(PO4)2.
c. It is quite likely that the two numbers from a and b will be different; obviously one of them must be wrong. The correct value is ALWAYS the smaller value and the reagent producing that value is the limiting reagent.
d. Convert moles Ba3(PO4)2 to grams. g = moles x molar mass

2. Using the coefficients as you did above, convert moles of the limiting reagent to moles HNO3. Then (H^+) = moles/2L and convert that to pH.

3. The nitrate ion will be the same concentration as the (H^+) in part 2.

To answer these questions, we need to calculate the number of moles, mass, and concentration of the involved substances in the given reaction.

1. Calculate the mass of Ba(PO4)2 formed:
We know from the balanced equation that 3 moles of Ba(NO3)2 react to form 1 mole of Ba3(PO4)2. Therefore, since we have 0.6 mol of Ba(NO3)2, we can calculate the number of moles of Ba3(PO4)2 formed as follows:

(0.6 mol Ba(NO3)2) x (1 mol Ba3(PO4)2 / 3 mol Ba(NO3)2) = 0.2 mol Ba3(PO4)2

To find the mass, we need to know the molar mass of Ba3(PO4)2, which can be calculated by summing the molar masses of each element:

Ba: 3(137.33 g/mol) = 411.99 g/mol
P: 2(30.97 g/mol) = 61.94 g/mol
O: 8(16 g/mol) = 128 g/mol

Total molar mass of Ba3(PO4)2 = 411.99 g/mol + 61.94 g/mol + 128 g/mol = 601.93 g/mol

Therefore, the mass of Ba3(PO4)2 formed is:

(0.2 mol Ba3(PO4)2) x (601.93 g/mol) = 120.39 g Ba3(PO4)2

2. Calculate the pH of HNO3:
Since HNO3 is a strong acid, it dissociates completely in water to give H+ ions. Since the concentration of HNO3 is not provided, we will assume it is the same as the initial number of moles used, which is 0.3 mol.

pH is calculated using the equation: pH = -log[H+]

So, pH = -log(0.3) = 0.522

3. Determine the concentration of the nitrate ion after the reaction reaches completion:
In the balanced equation, it is indicated that 6 moles of HNO3 form for every 3 moles of Ba(NO3)2. Therefore, we have:

(0.3 mol H3PO4) x (6 mol HNO3 / 2 mol H3PO4) = 0.9 mol HNO3

Since the initial volume of the solution is 2 L, the concentration of the nitrate ion is:

0.9 mol / 2 L = 0.45 mol/L

So, the concentration of the nitrate ion after the reaction reaches completion is 0.45 mol/L.