Asked by joshua
A mixture is prepared by adding 28.6 mL of 0.169 M Na3PO4 to 39.7 mL of 0.255 M Ca(NO3)2. What mass of Ca3(PO4)2 will be formed?
Answers
Answered by
Steve
You have:
.0286*.169 = 0.004833 moles of Na3PO4
.0397*.255 = 0.010124 moles of Ca(NO3)2
If the reaction is
2Na3PO4 + Ca(NO3)2 = 2NaNO3 + Ca(PO4)2
then you need 2 moles of Na for each mole of Ca.
Since there is extra Ca, you will wind up with
.09666 moles of Ca(PO4)2
Just multiply that by the mol mass of the compound.
.0286*.169 = 0.004833 moles of Na3PO4
.0397*.255 = 0.010124 moles of Ca(NO3)2
If the reaction is
2Na3PO4 + Ca(NO3)2 = 2NaNO3 + Ca(PO4)2
then you need 2 moles of Na for each mole of Ca.
Since there is extra Ca, you will wind up with
.09666 moles of Ca(PO4)2
Just multiply that by the mol mass of the compound.
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