In: Chemistry
A mixture contains 58.50 % BaCl2 and 41.50 % Na3PO4 by weight. When dissolved in water, the following double displacement reaction forms the barium phosphate precipitate. If you start with 6.00 grams of the mixture, how many grams of barium phosphate precipitate should form?
3 BaCl2 (aq) + 2 Na3PO4 (aq) --> 6 NaCl (aq) + Ba3(PO4)2 (s)
Use the following molar masses:
BaCl2 = 208.23 g/mol
Na3PO4 = 163.91 g/mol
NaCl = 58.44 g/mol
Ba3(PO4)2 = 601.93 g/mol
Enter your answer in grams, rounded to 3 significant figures. Do not include units.
Solution :-
Given data
Mass of mixture = 6.00g
% of BaCl2 =58.50%
%Na3PO4 = 41.50 %
Lets find the mass of each compound using the given percent
Mass of BaCl2 = 6.00 g mixture x 58.5 % BaCl2 / 100 % = 3.51g BaCl2
Mass of Na3PO4 = 6.00 g - 3.51 g = 2.49 g
Balanced reaction equation is as follows,
3 BaCl2 (aq) + 2 Na3PO4 (aq) --> 6 NaCl (aq) + Ba3(PO4)2 (s)
Now using the above balanced reaction equation lets find the mass of barium phosphate that can be formed using the mass of each reactant
Calculating using the mass of BaCl2
(3.51 g BaCl2 *1 mol /208.23 g) x ( 1 mol Ba3(PO4)2 / 3 mol BaCl2) x(601.93 g / 1 mol Ba3(PO4)2) = 3.38 g Ba3(PO4)2
Calculating using mass of Na3PO4
(2.49g Na3PO4 * 1mol / 163.91 g) x(1 mol Ba3(PO4)2/ 2 mol Na3PO4) x (601.93 g / 1mol Ba3(PO4)2) = 4.57 g Ba3(PO4)2
Since the BaCl2 gives the less mass of the product therefore it is limiting reactant and therefore the mass of Ba3(PO4)2 that can be formed is 3.38 g