6.12 moles of Na3PO4 experimentally produced 1001g of sodium chloride according to the equation below. What is the percent yield?

2Na3PO4 + 3 CaCl2 --> Ca3(PO4)2 + 6NaCl

155.5%

Explanation:

from the equation

2mol of Na3PO4 reacts to give 6mol of Nacl

therefore 6.12mol of Na3PO4 will give 18.36mol of Nacl

2mol Na3PO4 6mol of nacl

6.12mol NaPO4 6×6.12mol/2mol = 18.36mol of Nacl

mass of Nacl = mol × molar mass

where molar mass of Nacl= 23+12=35g/mol

mass of Nacl= 18.36 × 35= 642.6g of Nacl

percentage yield = actual yield/ theoretical yield ×100%

percentage yield= 1001g (given)/642.6g(used) ×100 = 155.5%

107%

Explanation

6.12 mol Na3Po4 | 6 mol NaCl | 58.44 g NaCl

= 1072g NaCl

                              | 2 mol Na3Po4 | 1 mol NaCl

1072g/1001g x 100= 107%

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