An ideal gas is stored in a container at constant volume. If the temperature (T) were increased to 3T, what would be the change
2 answers:
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The volume of O₂ produced: 84.6 L
<h3>Further explanation</h3>Given
7.93 mol of dinitrogen pentoxide
T = 48 + 273 = 321 K
P = 125 kPa = 1,23365 atm
Required
Volume of O₂
Solution
Decomposition reaction of dinitrogen pentoxide
2N₂O₅(g)→4NO₂(g)+O₂ (g)
From the equation, mol ratio N₂O₅ : O₂ = 2 : 1, so mol O₂ :
= 0.5 x mol N₂O₅
= 0.5 x 7.93
= 3.965 moles
The volume of O₂ :
<u>Answer:</u> The percent yield of the water is 31.98 %
<u>Explanation:</u>
To calculate the number of moles, we use the equation:
.....(1)
- <u>For methane:</u>
Given mass of methane = 6.58 g
Molar mass of methane = 16 g/mol
Putting values in equation 1, we get:
- <u>For oxygen gas:</u>
Given mass of oxygen gas = 14.4 g
Molar mass of oxygen gas = 32 g/mol
Putting values in equation 1, we get:
The chemical equation for the combustion of methane is:
By Stoichiometry of the reaction:
2 moles of oxygen gas reacts with 1 mole of methane
So, 0.45 moles of oxygen gas will react with = of methane
As, given amount of methane is more than the required amount. So, it is considered as an excess reagent.
Thus, oxygen gas is considered as a limiting reagent because it limits the formation of product.
By Stoichiometry of the reaction
2 moles of oxygen gas produces 2 moles of water
So, 0.45 moles of oxygen gas will produce = moles of water
- Now, calculating the mass of water from equation 1, we get:
Molar mass of water = 18 g/mol
Moles of water = 0.45 moles
Putting values in equation 1, we get:
- To calculate the percentage yield of water, we use the equation:
Experimental yield of water = 2.59 g
Theoretical yield of water = 8.1 g
Putting values in above equation, we get:
Hence, the percent yield of the water is 31.98 %