Question

For the reaction shown, calculate how many grams of oxygen form when each quantity of reactant completely reacts. 2 KClO3(s) → 2 KCl(s) + 3 O2(g) 2.72 g KClO3 0.361 g KClO3 83.6 kg KClO3 22.5 mg KClO3

Answer 1

1)

Molar mass of KClO3,

MM = 1*MM(K) + 1*MM(Cl) + 3*MM(O)

= 1*39.1 + 1*35.45 + 3*16.0

= 122.55 g/mol

mass of KClO3 = 2.72 g

molar mass of KClO3 = 122.55 g/mol

mol of KClO3 = (mass)/(molar mass)

= 2.72/122.55

= 0.0222 mol

From balanced chemical reaction, we see that

when 2 mol of KClO3 reacts, 3 mol of O2 is formed

mol of O2 formed = (3/2)* moles of KClO3

= (3/2)*0.0222

= 0.0333 mol

mass of O2 = number of mol * molar mass

= 0.0333*32

= 1.07 g

Answer: 1.07 g

2)

mass of KClO3 = 0.361 g

molar mass of KClO3 = 122.55 g/mol

mol of KClO3 = (mass)/(molar mass)

= 0.361/122.55

= 0.0029 mol

mol of O2 formed = (3/2)* moles of KClO3

= (3/2)*0.0029

= 0.0044 mol

mass of O2 = number of mol * molar mass

= 0.0044*32

= 0.141 g

Answer: 0.141 g

3)

mass of KClO3 = 83.6 Kg = 83600 g

molar mass of KClO3 = 122.55 g/mol

mol of KClO3 = (mass)/(molar mass)

= 83600/122.55

= 682.1705 mol

mol of O2 formed = (3/2)* moles of KClO3

= (3/2)*682.1705

= 1023.2558 mol

mass of O2 = number of mol * molar mass

= 1023.2558*32

= 32744 g

= 32.7 Kg

Answer: 32.7 Kg

4)

mass of KClO3 = 22.5 mg = 0.0225 g

molar mass of KClO3 = 122.55 g/mol

mol of KClO3 = (mass)/(molar mass)

= 0.0225/122.55

= 1.836*10^-4 mol

mol of O2 formed = (3/2)* moles of KClO3

= (3/2)*1.836*10^-4 mol

= 2.754*10^-4 mol

mass of O2 = number of mol * molar mass

= 2.754*10^-4 mol*32 g/mol

= 8.81*10^-3 g

= 8.81 mg

Answer: 8.81 mg