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7. What is the pressure of 1.00 mol of Ar gas that has a volume of 843.0 mL and a temperature of -86.0 °C? Given: V= 843 mL = 0.843 L Req: P T1= By Ideal Gas law n= P V = nRT R= Show your calculation. 8. The equation for the formation of ammonia gas (NH3) is as follows: N2(g) + 3H2(g) ? 2NH3(g) At 500 °C and 1.00 atm, 10.0 L of N2 gas are reacted to make ammonia. a. If the pressures and temperatures of H2 and NH3 were the same as those of N2, what volume of H2 would be needed to react with N2, and what volume of NH3 gas would be produced? Ideal gas law; PV = nRT T = 500 °C + 273 °C = 773 K Mole of N2(g) = PV / RT = 1.00 atm x 10.0 L / (0.0821 atm·L/mole·K) x 773 K = 0.158 mole Based on the balanced chemical equation: Moles of H2 = 0.158 moles N2 x 3 moles H2 / 1 mole N2 = 0.474 mole PV = nRT; V = nRT / P Vol. of H2 = Moles of NH3 = 0.158 moles N2 x 2 moles NH3 / 1 mole N2 = 0.316 mole Vol. of NH3 =

          7. What is the pressure of 1.00 mol of Ar gas that has a volume of 843.0 mL and a temperature of -86.0 °C?
Given: V= 843 mL = 0.843 L Req: P
T1= By Ideal Gas law
n= P V = nRT
R=
Show your calculation.

8. The equation for the formation of ammonia gas (NH3) is as follows:
N2(g) + 3H2(g) ? 2NH3(g)
At 500 °C and 1.00 atm, 10.0 L of N2 gas are reacted to make ammonia.
a. If the pressures and temperatures of H2 and NH3 were the same as those of N2, what volume of H2 would be needed to react with N2, and what volume of NH3 gas would be produced? Ideal gas law; PV = nRT T = 500 °C + 273 °C = 773 K
Mole of N2(g) = PV / RT = 1.00 atm x 10.0 L / (0.0821 atm·L/mole·K) x 773 K = 0.158 mole
Based on the balanced chemical equation:
Moles of H2 = 0.158 moles N2 x 3 moles H2 / 1 mole N2 = 0.474 mole PV = nRT; V = nRT / P
Vol. of H2 =
Moles of NH3 = 0.158 moles N2 x 2 moles NH3 / 1 mole N2 = 0.316 mole
Vol. of NH3 =

7. What is the pressure of 1.00 mol of Ar gas that has a volume of 843.0 mL and a temperature of -86.0 °C?
Given: V= 843 mL = 0.843 L Req: P
T1= By Ideal Gas law
n= P V = nRT
R=
Show your calculation.

8. The equation for the formation of ammonia gas (NH3) is as follows:
N2(g) + 3H2(g) ? 2NH3(g)
At 500 °C and 1.00 atm, 10.0 L of N2 gas are reacted to make ammonia.
a. If the pressures and temperatures of H2 and NH3 were the same as those of N2, what volume of H2 would be needed to react with N2, and what volume of NH3 gas would be produced? Ideal gas law; PV = nRT T = 500 °C + 273 °C = 773 K
Mole of N2(g) = PV / RT = 1.00 atm x 10.0 L / (0.0821 atm·L/mole·K) x 773 K = 0.158 mole
Based on the balanced chemical equation:
Moles of H2 = 0.158 moles N2 x 3 moles H2 / 1 mole N2 = 0.474 mole PV = nRT; V = nRT / P
Vol. of H2 =
Moles of NH3 = 0.158 moles N2 x 2 moles NH3 / 1 mole N2 = 0.316 mole
Vol. of NH3 =

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