Question

At elevated temperatures, sodium chlorate decomposes to produce sodium chloride and oxygen gas. A 0.8765-g sample of impure sodium chlorate was heated until the production of oxygen gas ceased. The oxygen gas collected over water occupied 57.2 mL at a temperature of 22ºC and a pressure of 734 torr. Calculate the mass percent of NaHCO₃ in the original sample. (At 22ºC the vapor pressure of water is 19.8 torr.)

Answer 1

Answer:

z = 18,46%

Explanation:

First we need to discover the mole number of the O2 produced so we use PV=nRT

734 torr * 57.2*10^-3 L = n 62.3637 * 295.15 K

n = $\frac{734 * 57.2*10^-3}{62.3637 * 295.15}$

n = $\frac{41,98}{18.406}$

n = 2,28 * 10^-3 mole

The reaction to the decomposition is:

2 NaClO3 --> 2 NaCl + 3 O2

2 mole NaClO3 - 3 moles O2

x - 2,28*10^-3 moles O2

X= 1,52*10^-3 moles of NaClO3

1 moles of NaClO3 - 106,44 g

1,52*10^-3 moles of NaClO3 - y

y = 0,1617 g

0,8765 g - 100%

0,1617 g - z

z = 18,46%