ignment/tak assignment-take Use the References to access insportant values if ne

Question

ignment/tak assignment-take Use the References to access insportant values if needed for this question For the following reaction, 15.7 grams of phosphorus (P) are allowed to react with 41.2 grams of chlorine gas. phosphorus (P) (s)+ chlorine (g)phosphorus trichloride ) What is the maximum amount of phosphorus trichloride that can be formed? grams What is the FORMULA for the limiting reagent? What amount of the excess reagent remains after the reaction is complete? grams Submit Answer Retry Entire Group 9 more group attempts remaining

Explanation / Answer

P4 (s) + 6 Cl2 (g) -----------> 4 PCl3 (l)

Mass of P4 = 15.7 g.

Molar mass of P4 = 4 * 31 = 124 g/mol

Moles of P4 = mass / molar mass = 15.7 / 124 = 0.127 mol

Mass of Cl2 = 41.2 g.

Molar mass of Cl2 = 71.0 g/mol

Moles of Cl2 = 41.2 / 71.0 = 0.580 mol

Fromt the balanced equation,

1 mol of P4 needs 6 mol of Cl2

then, 0.127 mol of P4 needs 6 * 0.127 = 0.762 mol of Cl2.

But we have only 0.580 mol of Cl2

Hence the limiting reagent = Cl2

From the balanced equation,

6 mol of Cl2 forms 4 mol of PCl3

Then,

0.580 mol of Cl2 forms 0.580 * 4 / 6 = 0.387 mol of PCl3

Mass of PCl3 = moles * molar mass = 0.387 * 137.5 = 53.2 g.

Moles of excess reagent remained = 0.127 - (0.580 * / 6) = 0.0303 mol

Mass of excess reagent remained = 0.0303 * 124 = 3.76 g.

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