Calculate the volume a solution required toof provide the following: (a) 2.14 g of sodium chloridefrom a 0.270 M solution, (b) 4
.30 g of ethanol from a1.50M solution, (c) 0.85 g of acetic acid (CH3COOH)from a 0.30 M solution.
1 answer:
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Answer:
Moles of potassium chlorate reacted = 0.2529 moles
The amount of oxygen gas collected will be 12.8675 g
Explanation:
(a)
We are given:
Vapor pressure of water = 17.5 mmHg
Total vapor pressure = 748 mmHg
Vapor pressure of Oxygen gas = Total vapor pressure - Vapor pressure of water = (748 - 17.5) mmHg = 730.5 mmHg
To calculate the amount of Oxygen gas collected, we use the equation given by ideal gas which follows:
where,
P = pressure of the gas = 730.5 mmHg
V = Volume of the gas = 9.49 L
T = Temperature of the gas =
R = Gas constant =
n = number of moles of oxygen gas = ?
Putting values in above equation, we get:
According to the reaction shown below as:-
3 moles of oxygen gas are produced when 2 moles of potassium chlorate undergoes reaction.
So,
0.3794 mol of oxygen gas are produced when moles of potassium chlorate undergoes reaction.
<u>Moles of potassium chlorate reacted = 0.2529 moles</u>
(b)
We are given:
Vapor pressure of water = 17.5 mmHg
Total vapor pressure = 753 mmHg
Vapor pressure of Oxygen gas = Total vapor pressure - Vapor pressure of water = (753 - 17.5) mmHg = 735.5 mmHg
To calculate the amount of Oxygen gas collected, we use the equation given by ideal gas which follows:
where,
P = pressure of the gas = 735.5 mmHg
V = Volume of the gas = 9.99 L
T = Temperature of the gas =
R = Gas constant =
n = number of moles of oxygen gas = ?
Putting values in above equation, we get:
Moles of Oxygen gas = 0.40211 moles
Molar mass of Oxygen gas = 32 g/mol
Putting values in above equation, we get:
<u>Hence, the amount of oxygen gas collected will be 12.8675 g</u>