In a precipitate reaction between BaCl2(aq) and Na3PO4(aq), 57 mL of 0.100 M BaCl2(aq) completely reacted...
In a precipitate reaction between BaCl2(aq) and Na3PO4(aq), 57 mL of 0.100 M BaCl2(aq) completely reacted with 25.7 mL of Na3PO4(aq). What was the molarity of Na3PO4(aq)?
3 BaCl2(aq) + 2 Na3PO4(aq) → Ba3(PO4)2(s) + 6 NaCl(aq)
Homework Answers
3 BaCl2(aq) + 2 Na3PO4(aq) → Ba3(PO4)2(s) + 6 NaCl(aq)
using the dilution law M1V1 /n1 = M2V2 /n2
where M1 = 0.100M , V1 = 0.057 L , n1 = 3 ===============> for BaCl2
M2 = ? , V2 = 0.0257 L , n2 = 2 =================> for Na3PO4
SUBSTITUTING THE VALUES
M2 = M1*V1*n2 / n1*V2
=> M2 = 0.100*0.057*2 / 3*0.0257
=> M2 = 0.1478M
answer => 0.1478 M of Na3PO4
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In a precipitate reaction between BaCl2(aq) and Na3PO4(aq), 57
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In a precipitate reaction between BaCl2(aq) and Na3PO4(aq), 56.7 mL of 0.100 M BaCl2(aq) completely reacted with 25.3 mL of Na3PO4(aq). What was the molarity of Na3PO4(aq)? 3 BaCl2(aq) + 2 Na3PO4(aq) → Ba3(PO4)2(s) + 6 NaCl(aq) Note: Insert only the numerical value of your answer (do not include the units or chemical in your answer).
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