Consider the reaction: 2H2O (l) ----> 2h2 (g) + O2 (g) What mass of H2O (MW=18.02)...

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Question

Consider the reaction: 2H2O (l) ----> 2h2 (g) + O2 (g) What mass of H2O (MW=18.02)...

Consider the reaction: 2H₂O (l) ----> 2h₂ (g) + O₂ (g)

What mass of H₂O (MW=18.02) is required to form 1.4 L of O₂ at a temperature of 30.0 degrees C and a pressure of 750 torr?

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Answer 1

Answer #1

Lets 1st calculate the mol of O2 formed.

Given:

P = 750 torr

= (750/760) atm

= 0.9868 atm

V = 1.4 L

T = 30.0 oC

= (30.0+273) K

= 303 K

find number of moles using:

P * V = n*R*T

0.9868 atm * 1.4 L = n * 0.08206 atm.L/mol.K * 303 K

n = 5.557*10^-2 mol

From reaction,

Mol of H2O reacted = 2*mol of O2 formed

= 2*5.557*10^-2 mol

= 0.1111 mol

Molar mass of H2O,

MM = 2*MM(H) + 1*MM(O)

= 2*1.008 + 1*16.0

= 18.016 g/mol

use:

mass of H2O,

m = number of mol * molar mass

= 0.1111 mol * 18.02 g/mol

= 2.002 g

Answer: 2.00 g

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Answer 2

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