What volume of oxygen gas, at 30.0ºC and 1.50 atm, is produced when 75.0g of water is electrolyzed? 2H2O --> 2H2 + O2

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Answer 1 · 2018-03-16T18:33:48

$34.55 L$ $"34.55 L"$ $(9.13 gal)$ $"(9.13 gal)"$

$2 H_{2} O - - - - \to 2 H_{2} + O_{2}$ $"2H_2 O -----> 2H_2 + O_2$

FROM DATA

Moles of WATER
$\frac{75 g}{18 g/mol} =$ $"75 g" / "18 g/mol" =$ $4.167 mol of H_{2} O$ $"4.167 mol of"" H_2O$

Mol of OXYGEN
$\frac{O_{2}}{2 H_{2} O} 4.167 =$ $O_2 / ("2 "H_2 O") 4.167=$ $2.083 mol of O_{2}$ $"2.083 mol of"" O_2$

CALCULATING OXYGEN VOLUME per IDEAL GAS LAW
(Relatively low pressure and temperature)

$V = \frac{nRT}{P}$ $V = "nRT"/"P"$

Where,
n: moles
R: Ideal gas constant, (0.0821 L. Atm/mol. L)
T: Temperature. (K)
P: Pressure, (Atm.)

Solving (Volume of Oxygen),

$V = \frac{(2.083)(00821)(273 + 30)}{1.5}$ $V = "(2.083)(00821)(273 + 30)"/"1.5"$ = $34.55 L$ $"34.55 L"$ $(9.13 gal)$ $"(9.13 gal)"$

What volume of oxygen gas, at 30.0ºC and 1.50 atm, is produced when 75.0g of water is electrolyzed? 2H﻿2﻿O --> 2H﻿2﻿ + O﻿2﻿