P4 + 10 Cl2 -----> 4 PCl5 Moles of P4 = 20 g / 123.9 g/mol => 0.1614 mol Moles o

Question

P4 + 10 Cl2 -----> 4 PCl5

Moles of P4 = 20 g / 123.9 g/mol => 0.1614 mol

Moles of Cl2 = 55 g / 70.9 g/mol => 0.7757 mol

1 mol of P4 needs 10 mol of Cl2. So

0.1614 mol of P4 would require 0.1614 *10 => 1.614 mol of Cl2

But available Cl2 is only 0.7757 mol. So Cl2 is limiting reactant.

Therefore amount of P4 that could react with available Cl2 is

0.7754 * (1/10) => 0.07757 moles of P4

1 equivalent of P4 can produce 4 equivalent of PCl5. So

0.07757 mol of P4 can produce 4*0.07757 => 0.3103 mol of PCl5

Mass of PCl5 = 0.3103 mol x 208.24 g/mol => 64.61 g of PCl5

Explanation / Answer

Hints 0 310 mole Submit My Answers Give Up Correct Part D What mass of PCls will be produced from the given masses of both reactants? Express your answer to three significant figures and include the appropriate units. Hints Value Units Submit My Answers Give Up

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