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if 3.95 mol of an ideal gas has a pressure of 2.33 atm and a volume of 60.87 l, what is the temperature of the sample in degrees
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Answer:- 1.24 g
Solution:- The balanced equation for the formation of carbonic acid by the reaction of carbon dioxide with water is:
From balanced equation, there is 1:1 mol ratio between carbon dioxide and carbonic acid. So, moles of carbonic acid will be equal to the moles of carbon dioxide used.
Moles of carbon dioxide can be calculated using ideal gas law equation as it's volume, temperature and pressure are given.
we need to convert mL to L, degree C to kelvin and kilopascals to atm.
= 0.495 L
25 + 273 = 298 K
= 1 atm
Ideal gas law equation is:
PV = nRT
We want to find out the n, so let's rearrange this:
R is the universal gas constant and it's value is .
Let's pug in the values in the equation and solve it for n.
n = 0.02 mol
To convert the moles to grams we multiply the moles by the molar mass of carbonic acid.
Molar mass of carbonic acid = 2(1.008)+12.01+3(16.00)
= 2.016+12.01+48.00
= 62.03 gram per mol (rounded to two decimal places)
Let's multiply the moles by molar mass:
= 1.24 g
So, the theoretical yield of carbonic acid is 1.24 g.
The mass formed in a chemical reaction is given by stoichiometry law in balanced chemical equation. The mass of silver produced in the reaction is 107. 9 g.
<h3>What is the stoichiometric law?</h3>The stoichiometric law states that in a balanced chemical equation, the stoichiometric coefficient describes the moles of each compound in a chemical reaction.
The balanced chemical equation for the reaction is:
From the balanced chemical equation, 1 moles of copper results in 2 moles of silver.
Moles in terms of mass can be given as:
Thus, moles of 31. 75 grams copper is:
The moles of Ag produced can be given from stoichiometric law as:
The moles of Ag produced is 1 mol.
The mass of a mole of element is equivalent to the molar mass. Thus, the mass of Ag produced in the reaction is 107. 9 g. Hence, option B is correct.
Learn more about stoichiometric law, here: