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Home Q & A Board Science-Related Homework Help Chemistry (Moderator: Laser_3)
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SMOKEY2112 SMOKEY2112
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11 years ago
Balance the equation and find which is the limiting reactant when 50 g of each reactant is used.  How many grams of Ba3(PO4)2 solide is produced?
Ba(OH)2 (aqueous)+ H3PO4(aqueous)-------->H2O (liquid) + Ba3(PO4)2(solid)
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wrote...
#1
11 years ago
3 Ba(OH)2 + 2 H3PO4 = Ba3(PO4)2 + 6 H2O

H3PO4 is the limiting reagent.  307.12 g are formed.
wrote...
#2
11 years ago
3 Ba(OH)2 + 2 H3PO4 = Ba3(PO4)2 + 6 H2O
moles Ba(OH)2 = 50 g / 171.34 g/mol=0.292

moles H3PO4 = 50 g / 90.0 g/mol=0.556

the ratio between Ba(OH)2 and H3PO4 is 3 : 2

moles H3PO4 needed = 0.292 x 2/3 =0.195
we have 0.556 mol of H3PO4 so it is in excess and Ba(OH)2 is the limitng reactant

moles barium phosphate = 0.292 / 3=0.0973

mass Ba3(PO4)2 = 0.0973 mol x 601.84 g/mol=58.6 g
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