Question

Calculate the number of grams of potassium chlorate required to produce 1.22L of oxygen gas at 17.0 Celcius and 793 Torr, given the following unbalanced chemical equation KCIO3(s) -> KCI(s) + O2(g)

A: 0.225g

B: 74.6 g

C: 4.37 g

D: 56.7 kg

Answer 1

The balanced equation is

2 KClO3 ------> 2 KCl + 3 O₂

PV = nRT

where, P = 793 torr = 1.04 atm

V = volume = 1.22 L

n = number of moles = ?

R = gas constant

T = temperature = 17.0 + 273 = 290 K

1.04*1.22 = n * 0.0821*290

1.27 = n * 23.8

n = 1.27 / 23.8 = 0.0534 mol

number of moles of oxygen = 0.0534 mol

From the balanced equation we can say that

3 mole of O2 produced by 2 mole of KClO3 so

0.0534 mole of O2 will be produced by

= 0.0534 mole of O2 *(2 mole of KClO3 / 3 mole of O2)

= 0.0356 mole of KClO3

mass of 1 mole of KClO3 = 122.55 g

so the mass of 0.0356 mole of KClO3 = 4.37 g

Therefore, the mass of KClO3 required would be 4.37 g