Question

Calculate the following. Show your work. Circle or place a box around your final answer. A frequently used method for preparing oxygen in the laboratory is by the thermal decomposition of Potassium Chlorate (KCIOs). Use the equation provided to answer the questions below. Given: Density of Oz - 1.428 g/L KCIO3(s) KCI (s) + O2 (g) 1. Balance the above equation (Quiz 10) 10 pts 2. What volume of O, can be made from 5.00 x 10 moles of KCIO,? (Quiz 11) 10 pts 13. How many grams of KCIO, must react to form 42.0 mL 02? (Quiz 12) 10 pts 4. How many ml of O, will form from 55.2 g KCIO,? (Quiz 13) 10 pts

Answer 1

Ans 1

The balanced reaction

2KClO3 (s) = 2KCl (s) + 3O2 (g)

Ans 2

Moles of KClO3 = 5*10^-2 mol

From the stoichiometry of the reaction

2 moles of KClO3 produces = 3 mol O2

5*10^-2 mol of KClO3 produces = 5*10^-2 * 3/2

= 0.075 mol O2

Mass of O2 produced = moles x molecular weight

= 0.075 mol x 32 g/mol

= 2.40 g

Volume of O2 = mass/density

= (2.40 g) / (1.428 g/mL)

= 1.68 mL

Ans 3

Volume of O2 = 42 mL

Mass of O2 = volume x density

= 42 mL x 1.428 g/mL

= 59.976 g

Moles of O2 = mass/molecular weight

= (59.976 g) / (32 g/mol)

= 1.874 mol

From the stoichiometry of the reaction

Moles of KClO3 reacted = 1.874 mol * 2/3

= 1.25 mol

Mass of KClO3 reacted = moles x molecular weight

= 1.25 mol x 122.55 g/mol

= 153.12 g

Ans 4

Mass of KClO3 = 55.2 g

Moles of KClO3 = mass/molecular weight

= (55.2g) / (122.55 g/mol)

= 0.450 mol

Moles of O2 formed = 0.450 mol x 3/2 = 0.676 mol

Mass of O2 formed = moles x molecular weight

= 0.676 mol x 32 g/mol

= 21.62 g

Volume of O2 = mass/density

= ( 21.62 g) / (1.428 g/mL)

= 15.14 mL