Question

A mixture of KCl and KCl0; weighing 1. 34 grams was heated; the dry 02 generated occupied 143 mL at STP. What percent by mass of the original mixture was KCIO, which decomposes as follows: 2KCIO, (s) + 2KCi (s) + 302 (g)

Answer 1

1st find mol of O2

Given:

P = 1.0 atm

V = 143.0 mL

= (143.0/1000) L

= 0.143 L

T = 273.0 K

find number of moles using:

P * V = n*R*T

1 atm * 0.143 L = n * 0.08206 atm.L/mol.K * 273 K

n = 6.383*10^-3 mol

From reaction,

Mol of KClO reacted = (2/3)* mol of O2 formed

= (2/3)*6.383*10^-3 mol

= 4.26*10^-3 mol

Molar mass of KClO,

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

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

= 90.55 g/mol

use:

mass of KClO,

m = number of mol * molar mass

= 4.26*10^-3 mol * 90.55 g/mol

= 0.3857 g

Now use:

Mass % of KClO = mass of KClO * 100 / mass of mixture

= 0.3857 * 100 / 1.34

= 28.8 %

Answer: 28.8 %