1.1296 grams of O2 is produced.
Explanation:
n = number of mole
[tex]n = \frac{PV}{RT}\\[/tex]
[tex]P = 1.015 - (\frac{17.5}{760})\\ P =0.992 atm[/tex]
[tex]T = 273.15 + 20.0 = 293.15 K[/tex]
R = 0.082057
V = 0.856 L
Therefore,
[tex]n = \frac{0.992 \times 0.856}{0.082057 \times293.15}[/tex]
n = 0.0353 mole
g = 0.0353 mole x 32.0 g/mole = 1.1296 g of O2
The mass of O2 produced is 1.126g grams
Since the decomposition of KClO3 when 856 mL of O2 is collected over water at 20.0°C and 1.015 atm. The vapor pressure of water at 20.0°C is 17.5 torr.
Now we know that
n = PV/RT
Here P should be = 1.015-(17.5/760)
= 0.992 atm
Now the T should be
= 273.15 + 20.0
= 293.15 K
Since R = 0.082057 and V = 0.856 L
Now the mass is
= (0.992*0.856) / (0.82057 * 293.15)
= 0.0353 mole
Now g should be
= 0.0353 mole x 32.0 g/mole
= 1.1296 g of O2
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