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.
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?
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