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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 =
n =
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%