Question

c. (10 pts.) Using the equation: 3 S-Cl2 (aq) + 2 Li3PO4 (aq) → Sr3(PO4)2 (8) + 6 LICI (aq) If you are given 5.80 g of SrCl2
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Answer #1

Starting material/product

Molecular weight

Amount taken

moles

Equivalent

SrCl2

158.53

5.80 gm

0.0365 mol

1.729

Li3PO4

115.79

2.45 gm

0.0211 mol

1.00

Sr3(PO4)2

452.8027

Li3PO4 is the limiting reagent in the reaction

115.79 gm of Li3PO4 gives 452.8027 gm of Product

Therefore 2.45 gm of Li3PO4 will give x gm of product

Theoretical yield of the reaction = (2.45*452.8027)/115.79 = 9.58 gm

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c. (10 pts.) Using the equation: 3 S-Cl2 (aq) + 2 Li3PO4 (aq) → Sr3(PO4)2 (8)...
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