Answer: There are 7.4 moles of helium gas present in a 1.85 liter container at the same temperature and pressure.
Explanation:
Given: = 2.25 L,
= 9.0 mol
= 1.85 L,
= ?
Formula used to calculate the moles of helium are as follows.
Substitute the values into above formula as follows.
Thus, we can conclude that there are 7.4 moles of helium gas present in a 1.85 liter container at the same temperature and pressure.
<u>Answer:</u> The mass of nitrogen gas reacted to produce given amount of energy is 5.99 grams.
<u>Explanation:</u>
The given chemical reaction follows:
We know that:
Molar mass of nitrogen gas = 28 g/mol
We are given:
Enthalpy change of the reaction = 14.2 kJ
To calculate the mass of nitrogen gas reacted, we use unitary method:
When enthalpy change of the reaction is 66.4 kJ, the mass of nitrogen gas reacted is 28 grams.
So, when enthalpy change of the reaction is 14.2 kJ, the mass of nitrogen gas reacted will be =
Hence, the mass of nitrogen gas reacted to produce given amount of energy is 5.99 grams.
The volume of oxygen at STP required would be 252.0 mL.
<h3>Stoichiometic problem</h3>The equation for the complete combustion of C2H2 is as below:
The mole ratio of C2H2 to O2 is 2:5.
1 mole of a gas at STP is 22.4 L.
At STP, 100.50 mL of C2H2 will be:
100.50 x 1/22400 = 0.0045 mole
Equivalent mole of O2 according to the balanced equation = 5/2 x 0.0045 = 0.01125 moles
0.01125 moles of O2 at STP = 0.01125 x 22400 = 252.0 mL
Thus, 252.0 mL of O2 gas will be required at STP.
More on stoichiometric problems can be found here: brainly.com/question/14465605
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