In a single displacement reaction between Sodium Phosphate and Barium, how much of each product (in grams) will be formed from 10.0 grams of Barium? Show your work.

14.6 g of barium phosphate

3.35 g of sodium metal

Explanation:

2Na3PO4(aq) + 3Ba(s) > Ba3(PO4)2(aq) + 6Na(s)

The first step in any such reaction is to but down the balanced reaction equation according to the stoichiometry of the reaction.

The two products formed are barium phosphate and sodium metal.

Number of moles of barium corresponding to 10.0g of barium = mass of barium/ molar mass of barium

Molar mass of barium = 137.327 g

Number of moles of barium = 10/137.327

Number of moles of barium = 0.0728 moles

For barium phosphate;

3 moles of barium yields 1 mole of barium phosphate

0.0728 moles yields 0.0728 moles × 1/3 = 0.0243 moles of barium phosphate

Molar mass of barium phosphate = 601.93 g/mol

Therefore mass of barium phosphate = 0.0243 moles × 601.93 g/mol = 14.6 g of barium phosphate

For sodium metal

3 moles of barium yields 6 moles of sodium metal

0.0728 moles of barium yields 0.0728 × 6 / 3 = 0.1456 moles of sodium

Molar mass of sodium metal= 23 gmol-1

Mass of sodium metal= 0.1456g × 23 gmol-1 = 3.35 g of sodium metal

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