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# You mixed 10.0 mL of a 0.85 M Na3PO4 solution with 8.5 mL of a 1.2 M ZnCl2 solution. ?

Chemical Equation:

3 ZnCl2 + 2 Na3PO4 > Zn3(PO4)2 + 6 NaCl

What is the theoretical yield in moles of Zn3(PO4)2 based on Na3PO4?

What is the theoretical yield in moles of Zn3(PO4)2 based on ZnCl2?

Based on the limiting reagent, how many grams of Zn3(PO4)2 should be produced.

Please list the steps. Thank you!

### 2 Answers

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- Roger the MoleLv 77 months agoFavourite answer
(0.0100 L) x (0.85 mol/L Na3PO4) x (1 mol Zn3(PO4)2 / 2 mol Na3PO4) = 0.00425 mol Zn3(PO4)2 in theory

(0.0085 L) x (1.2 mol/L ZnCl2) x (1 mol Zn3(PO4)2 / 3 mol ZnCl2) = 0.00340 mol Zn3(PO4)2 in theory

(0.00340 mol Zn3(PO4)2) x (386.0827 g Zn3(PO4)2/mol) = 1.31268 g = 1.3 g Zn3(PO4)2

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- 7 months ago
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