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VladimirAG [237]
3 years ago
7

The Haber Process synthesizes ammonia at elevated temperatures and pressures. Suppose you combine 1580 L of nitrogen gas and 351

0 L of hydrogen gas at STP, heat the mixture to run the reaction, then separate the ammonia from the reaction mixture. What volume of reactant, measured at STP, is left over? Assume the reaction goes to completion. N₂ (g) + 3 H₂ (g) → 2 NH₃ (g)
Chemistry
1 answer:
ikadub [295]3 years ago
8 0

Answer : The volume of reactant measured at STP left over is 409.9 L

Explanation :

First we have to calculate the moles of N_2 and H_2 by using ideal gas equation.

<u>For N_2 :</u>

PV_{N_2}=n_{N_2}RT

where,

P = Pressure of gas at STP = 1 atm

V = Volume of N_2 gas = 1580 L

n = number of moles N_2 = ?

R = Gas constant = 0.0821L.atm/mol.K

T = Temperature of gas at STP = 273 K

Putting values in above equation, we get:

1atm\times 1580L=n_{N_2}\times (0.0821L.atm/mol.K)\times 273K

n_{N_2}=70.49mole

<u>For H_2 :</u>

PV_{H_2}=n_{H_2}RT

where,

P = Pressure of gas at STP = 1 atm

V = Volume of H_2 gas = 3510 L

n = number of moles H_2 = ?

R = Gas constant = 0.0821L.atm/mol.K

T = Temperature of gas at STP = 273 K

Putting values in above equation, we get:

1atm\times 3510L=n_{H_2}\times (0.0821L.atm/mol.K)\times 273K

n_{H_2}=156.6mole

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

N_2(g)+3H_2(g)\rightarrow 2NH_3(g)

From the balanced reaction we conclude that

As, 3 mole of H_2 react with 1 mole of N_2

So, 156.6 moles of H_2 react with \frac{156.6}{3}\times 1=52.2 moles of N_2

From this we conclude that, N_2 is an excess reagent because the given moles are greater than the required moles and H_2 is a limiting reagent and it limits the formation of product.

Now we have to calculate the excess moles of N_2 reactant (unreacted gas).

Excess moles of N_2 reactant = 70.49 - 52.2 = 18.29 moles

Now we have to calculate the volume of reactant, measured at STP, is left over.

PV=nRT

where,

P = Pressure of gas at STP = 1 atm

V = Volume of gas = ?

n = number of moles of unreacted gas = 18.29 moles

R = Gas constant = 0.0821L.atm/mol.K

T = Temperature of gas at STP = 273 K

Putting values in above equation, we get:

1atm\times V=18.29mole\times (0.0821L.atm/mol.K)\times 273K

V=409.9L

Therefore, the volume of reactant measured at STP left over is 409.9 L

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