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Sagot :
Answer:
240.17 g Ba3(PO4)2
Explanation:
1. Determine the limiting reactant.
2H3PO4 + 3Ba(OH)2 --> Ba3(PO4)2 + 6H2O
moles H3PO4 = M x V = 3 x 0.286 = .858 moles H3PO4
moles Ba(OH)2 = M x V = 1.4 x 0.855 = 1.197 moles Ba(OH)2
ratio Ba(OH)2 : H3PO4 = 1.197: .858 = 1.39: 1
stoichiometric ratio Ba(OH)2 : H3PO4 = 3:2
Ba(OH)2is the limiting reactant
MM Ba3(PO4)2 = 601.92 g/mol
g Ba3(PO4)2 = moles Ba(OH)2 x(1 mol Ba3(PO4)2/3 moles Ba(OH)2) x (MM Ba3(PO4)2/ 1mol Ba3(PO4)2) = 1.197 x 1/3 x 601.92 = 240.17 g Ba3(PO4)2
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