Asked by eng
When a reaction mixture with a total volume of 1750 mL that is 0.0263 M aqueous K+ was stoichiometrically produced as per the balanced equation, what volume (mL) of 0.0326 M aqueous K3PO4 was required?
3 BaBr2(aq) + 2 K3PO4(aq) → Ba3(PO4)2(s) + 6 KBr(aq)
3 BaBr2(aq) + 2 K3PO4(aq) → Ba3(PO4)2(s) + 6 KBr(aq)
Answers
Answered by
DrBob222
I would approach the problem this way.
0.0263 M x 0.1750 L = ?? moles K^+.
It will take ??moles K^+ x (2 mols K3PO4/6 moles K^+) = xx moles K3PO4.
Then M = moles/L. You know moles and M, solve for liters and multiply by 1000.
Actually, it is easier to work in millimoles.
0.263 x 1750/3 = millimoles K3PO4.
Then millimoles/mL = M and you solve for mL directly. Check my thinking (but don't invert the multiplier. ;-).
0.0263 M x 0.1750 L = ?? moles K^+.
It will take ??moles K^+ x (2 mols K3PO4/6 moles K^+) = xx moles K3PO4.
Then M = moles/L. You know moles and M, solve for liters and multiply by 1000.
Actually, it is easier to work in millimoles.
0.263 x 1750/3 = millimoles K3PO4.
Then millimoles/mL = M and you solve for mL directly. Check my thinking (but don't invert the multiplier. ;-).
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