Respuesta :
Answer: The mass of water required is 1.79 grams
Explanation:
To calculate the amount of oxygen gas, we use the equation given by ideal gas which follows:
[tex]PV=nRT[/tex]
where,
P = pressure of the gas = 0.926 atm
V = Volume of the gas = 1.3 L
T = Temperature of the gas = 295 K
R = Gas constant = [tex]0.0821\text{ L. atm }mol^{-1}K^{-1}[/tex]
n = number of moles of oxygen gas = ?
Putting values in above equation, we get:
[tex]0.926atm\times 1.3L=n\times 0.0821\text{ L. atm }mol^{-1}K^{-1}\times 295K\\\\n=\frac{0.926\times 1.3}{0.0821\times 295}=0.0497mol[/tex]
For the given chemical equation:
[tex]2H_2O(l)\rightarrow 2H_2(g)+O_2(g)[/tex]
By Stoichiometry of the reaction:
1 mole of oxygen gas is produced from 2 moles of water
So, 0.0497 moles of oxygen gas will be produced by = [tex]\frac{2}[1}\times 0.0497=0.0994mol[/tex] of water
To calculate the number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]
Moles of water = 0.0994 moles
Molar mass of water = 18 g/mol
Putting values in above equation, we get:
[tex]0.0994mol=\frac{\text{Mass of water}}{18g/mol}\\\\\text{Mass of water}=(0.0994mol\times 18g/mol)=1.79g[/tex]
Hence, the mass of water required is 1.79 grams
The mass of required to form 1.3 L of at a temperature of 295 K and a pressure of 0.926 atm is 1.79 g.
The number of moles of oxygen involved in the reaction can be calculated by using the ideal gas equation as follows:-
.....(1)
Where;
P = pressure (atm)=0.926 atm
V = volume (L)=1.3 L
n = number of moles (mol)
R = gas law constant (0.0821 Latm/molK)
T = temperature (K)=295 K
Substitute all the above values in the equation (1) as follows:-
[tex]n =\frac{0.926\times 1.3}{0.0821\times 295}\\\\=\frac{1.2308}{24.2195} \\\\=0.0497\ mol[/tex]
According to this reaction; [tex]2H_2O(l) \rightarrow 2H_2(g) + O_2(g)[/tex]
2 moles of [tex]H_2O[/tex] produce 1 mole of [tex]O_2[/tex]
0.0497 moles of will be produced when 0.0497 × 2 = 0.0994 moles of .
Therefore, the number of moles of [tex]H_2O[/tex] required to form 1.3 L of [tex]O_2[/tex] at a temperature of 295 K and a pressure of 0.926 atm is 0.0994 moles.
The mass of water can be calculated by using the equation as follows:-
[tex]Number \ of\ moles=\frac{Mass}{Molar\ mass}[/tex]
Substitute 0.0994 moles for number of moles and 18 g/mol for molar mass of water in the above equation as follows:-
[tex]0.0994\ mol=\frac{Mass}{18\ g/mol} \\\\Mass=0.0994\ mol\times18\ g/mol\\\\=1.79\ g[/tex]
Hence, the mass of [tex]H_2O[/tex] required to form 1.3 L of [tex]O_2[/tex] at a temperature of 295 K is 1.79 g.
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