Which type of pollution is created by nuclear power plants?

Explanation: PV=nRT, P is the pressure of gas, V is the volume it occupies n is the number of moles of gas present in the sample, R is the universal gas constant which is equal to 0.0821 atm L/mol K and T is the absolute temperature of the gas

8 0

3 years ago

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

ikadub [295]

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.

$PV_{N_2}=n_{N_2}RT$

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:

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

$n_{N_2}=70.49mole$

$PV_{H_2}=n_{H_2}RT$

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

8 0

2 years ago