Given the balanced equation N2(g)+3H2(g)==>2NH3
Calculate:
a)the volume H2 that reacts with 12L of N2
b)the volume of NH3 produced Iron 4L ofN2
c)the volume of N2 and H2 to produced 60L of NH3
Assume that all volume measurements are made under identical conditions
Consider a balloon that has a volume V. It contains n moles of gas, it has an internal pressure of P, and its temperature is T. If the balloon is heated to a temperature of 15.5T while it is placed under a high pressure of 15.5P, how does the volume of the balloon change?
12 L N2 x (3 mol H2/1 mol N2) = 12 x 3/1 = ?
The others are done the same way. Use the coefficients to convert one something to any other something. That's what they tell you. They aren't there to look pretty.
Consider a balloon that has a volume V. It contains n moles of gas, it has an internal pressure of P, and its temperature is T. If the balloon is heated to a temperature of 15.5T while it is placed under a high pressure of 15.5P, how does the volume of the balloon change?
no idea
it stays the same
To calculate the volumes of gases involved in a chemical reaction, we need to apply the principles of stoichiometry and the ideal gas law.
a) First, determine the molar ratio of H2 to N2 in the balanced equation. The equation shows that 3 moles of H2 react with 1 mole of N2. Therefore, the molar ratio is 3:1.
Next, calculate the number of moles of N2 in 12L. To do this, we need to know the molar volume of a gas at the given conditions. Let's assume the molar volume is 22.4 L/mol at standard temperature and pressure (STP).
Number of moles of N2 = volume of N2 / molar volume at STP
= 12L / 22.4 L/mol
= 0.536 moles of N2
Using the molar ratio from the balanced equation, we can determine the number of moles of H2 reacting with N2:
Number of moles of H2 = Number of moles of N2 × (3 moles of H2 / 1 mole of N2)
= 0.536 moles × 3/1
= 1.61 moles of H2
Now, we can calculate the volume of H2 using the ideal gas law:
PV = nRT
Let's assume the temperature (T) and pressure (P) are constant. Assuming ideal gas behavior, we can combine the ideal gas law equation and solve for the volume (V):
V = nRT / P
Remember that we are given the number of moles of H2, n = 1.61 moles, and all measurements are made under identical conditions.
b) Calculate the volume of NH3 produced from 4 L of N2:
First, determine the number of moles of N2:
Number of moles of N2 = volume of N2 / molar volume at STP
= 4 L / 22.4 L/mol
= 0.179 moles of N2
Using the molar ratio from the balanced equation, we can determine the number of moles of NH3 produced:
Number of moles of NH3 = Number of moles of N2 × (2 moles of NH3 / 1 mole of N2)
= 0.179 moles × 2/1
= 0.358 moles of NH3
Now, we can calculate the volume of NH3 using the ideal gas law. Assume the temperature and pressure are constant:
V = nRT / P
Given the number of moles of NH3, n = 0.358 moles, and all measurements are made under identical conditions.
c) To determine the volume of N2 and H2 needed to produce 60 L of NH3:
First, calculate the number of moles of NH3 using its volume and assumptions:
Number of moles of NH3 = volume of NH3 / molar volume at STP
= 60 L / 22.4 L/mol
= 2.68 moles of NH3
Using the molar ratio from the balanced equation, we can determine the number of moles of N2 required:
Number of moles of N2 = Number of moles of NH3 × (1 mole of N2 / 2 moles of NH3)
= 2.68 moles × 1/2
= 1.34 moles of N2
Similarly, calculate the moles of H2:
Number of moles of H2 = Number of moles of N2 × (3 moles of H2 / 1 mole of N2)
= 1.34 moles × 3/1
= 4.02 moles of H2
Finally, using the ideal gas law and assumptions:
V = nRT / P
Given the number of moles of N2 and H2, n = 1.34 moles and 4.02 moles, respectively, and all measurements are made under identical conditions.