Question 1 If 2.01 moles of nitrogen reacts with 4.96 moles of hydrogen, what is

Question

Question 1 If 2.01 moles of nitrogen reacts with 4.96 moles of hydrogen, what is the maximum volume of ammonia (in L) that can be produced at a pressure of 1.03 atm and 347 K. N2(g) + 3 H2(g) 2 NH3(g)

Question 2 15.9 g of butane (58.12 g/mol) undergoes combustion according to the following equation. What pressure of carbon dioxide in atm is produced at 309 K in a 1.71 L flask.

2 C4H10(g) + 13 O2 (g) 8 CO2 (g) + 10 H2O (g)

Question 3 An argon atom has a mass 10 times greater than a Helium atom. How many times faster is the helium atom's velocity?

Question 4 Which of the following will influence the speed at which a gas leaks out of a balloon? (select all that apply)

Size of the individual gas molecule

Mass of the individual gas molecule

Temperature of the gas molecule

Question 5 When measuring the volume and pressure of a real gas, are they expected to be larger or smaller than the values based on the ideal gas equation? Choose all correct answers.

Volume of real gas will be greater than predicted

Pressure of real gass wil be greater than predicted

Volume of the real gass will be lesser than predicted

Pressure of real gas will be lesser than predicted

Question 6 Under which of the following conditions is a gas expected to behave most like an ideal gas? Choose all that apply

Higher Temperature

Lower Temperature

Higher Pressure

Lower Pressure

Explanation / Answer

I am allowed to answer only 1 question at a time

1.
N2(g) + 3 H2(g) 2 NH3(g)
2.01 mol of N2 will require 3*2.01 = 6.03 mol of H2
clearly, H2 is limiting reagent since we have only 4.96 mol of H2

3 mol of H2 gives 2 mol of NH3
So,
number of moles of NH3 formed = (2/3)*4.96 mol = 3.31 mol

To calculate volume use:
P*V = n*R*T
1.03*V = 3.31*0.0821*347
V = 91.46 L
Answer: 91.46 L

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