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BaCl2+Na3PO4-->NaCl+Ba3 (PO4)2If 3.54g of Ba3 (PO4)2are produce by the reaction given how many grams of the starting material BCl2 were used?a) 3.680 gb) 1.23 gc) 3.68 gd) 4.16 g

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Answer:

1.23 g

Explanation:

Given that the mass of barium phosphate formed= 3.54 g

Molar mass of Ba3 (PO4)2 = 602 g/moles

Number of moles of Ba3 (PO4)2= 3.54g/602 g/moles = 0.0059 moles

Since 1 mole of BaCl2 forms 1 mole of Ba3 (PO4)2

Hence, 0.0059 moles of BaCl2 yields 0.0059 moles of Ba3 (PO4)2

Hence mass of BaCl2 reacted= mass × molar mass= 1.23 g

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