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The phosphorus pentoxide used to produce phosphoric acid for cola soft drinks is prepared by burning phosphorus in oxygen.

(a) What is the limiting reactant when 0.200 mol of P4 and 0.200 mol of O2 react according to

P4 + 5O2⟶P4O10

(b) Calculate the percent yield if 10.0 g of P4O10is isolated from the reaction.

Short Answer

Expert verified

a) Limiting reactant is O2

b) Percent yield will be 88 % of P4O10

Step by step solution

01

Determine limiting reactant

P4 + 5O2 ⟶P4 O10

0.200 0.200

mol mol

1 Mol of P4 gives 1moles of O2. Therefore,

(0.200 mol P4 x 5 mol of O2)/ 1 mol P4

= 1 mol of O2

Here, I have 0.200 mol of O2 but I need 1 mol of O2. Means 0.200 will get used up in the reaction. That’s why O2 is the limiting reactant.

02

Determine percentage yield

% yield = actual yield x 100

theoretical yield

Actual yield = 10 g ------given

Molar mass of P4O10 = 283.88

Theoretical yield = 0.200 mol O2 x 1 mol P4O10 x 283.88g P4O10

5 mol O2 1 mol P4O10

= 11.355 g P4O10

% yield = actual yield x 100

theoretical yield

\(\begin{aligned}{} &= \frac{{10.0}}{{11.355}} \times 100\\ &= 88\% \,\,{P_4}{O_{10}}\end{aligned}\)

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Most popular questions from this chapter

How many moles of Ca(OH)2 are required to react with 1.36 mol of H3PO4 to produce Ca3(PO4)2 according to the equation \(3Ca{\left( {OH} \right)_2} + 2{H_3}P{O_4} \to C{a_3}{\left( {P{O_4}} \right)_2} + 6{H_2}O\)?

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