Question

1. phosphoric acid can be prepared from elemental phosphorus, oxygen, and water by the following series of reactions. be sure to balance the following reactions.

what volume of acid is produced from reacting 1.250 grams of phosphorus with sufficient oxygen and water? the specific gravity of phosphoric acid is 1.18.

1. potassium chlorate decomposes on heating to form potassium chloride and oxygen.

a. write the balanced equation for the reaction.

b. if 4.904 grams of potassium chlorate, with a 75.00% purity, is heated, what mass of oxygen may be produced?

c. 1.208 grams of oxygen are experimentally obtained. what is the percent yield?

Explanation:

Step1: Write the balanced chemical equation

$$2KClO_3\stackrel{\Delta}{=\!=\!=}2KCl + 3O_2\uparrow$$

Step2: Calculate the mass of pure \(KClO_3\)

The mass of pure \(KClO_3\) is \(m = 4.904\times75.00\%= 3.678g\)

Step3: Calculate the molar mass of \(KClO_3\)

The molar mass of \(KClO_3\) is \(M = 39 + 35.5+3\times16=122.5g/mol\)

Step4: Calculate the number of moles of \(KClO_3\)

The number of moles of \(KClO_3\) is \(n=\frac{m}{M}=\frac{3.678}{122.5}=0.03mol\)

Step5: Calculate the number of moles of \(O_2\) according to the stoichiometry

From the balanced equation \(2KClO_3\sim3O_2\), the number of moles of \(O_2\) is \(n(O_2)=\frac{3}{2}n(KClO_3)=\frac{3}{2}\times0.03 = 0.045mol\)

Step6: Calculate the mass of \(O_2\)

The molar mass of \(O_2\) is \(M(O_2)=32g/mol\), and the mass of \(O_2\) is \(m(O_2)=n(O_2)\times M(O_2)=0.045\times32 = 1.44g\)

Step7: Calculate the percent yield

The percent yield \(\eta=\frac{actual\ mass}{theoretical\ mass}\times100\%=\frac{1.208}{1.44}\times100\%\approx83.9\%\)

Answer:

a. \(2KClO_3\stackrel{\Delta}{=\!=\!=}2KCl + 3O_2\uparrow\)
b. \(1.44g\)
c. \(83.9\%\)