Question

Cavendish prepared hydrogen in 1766 by the novel method of passing steam through a red-hot gun barrel: 4H2 O(g) + 3Fe(s) ⟶ Fe3 O4 (s) + 4H2 (g) (a) Outline the steps necessary to answer the following question: What volume of H2 at a pressure of 745 torr and a temperature of 20 °C can be prepared from the reaction of 15.O g of H2O? (b) Answer the question.

Answer 1

Answer : The volume of
`H_2` will be, 0.2690 L

Solution :

(a) Steps involved for this problem are :

First we have to calculate the moles of
`H_2O`.

Now we have to calculate the volume of hydrogen gas by using the ideal gas equation.

(b) First we have to calculate the moles of
`H_2O`.

`\text{Moles of }H_2O=\frac{\text{Mass of }H_2O}{\text{Molar mass of }H_2O}=(15g)/(18g/mole)=0.833moles`

The balanced chemical reaction is,

`4H_2O(g)+3Fe(s)\rightarrow Fe_3O_4(s)+4H_2(g)`

From the balanced chemical reaction, we conclude that the moles of hydrogen is equal to the moles of water.

Thus, the moles of hydrogen gas = 0.833 mole

Now we have to calculate the volume of hydrogen gas.

Using ideal gas equation,

`PV=nRT`

where,

n = number of moles of gas = 0.833 mole

P = pressure of the gas =
`745torr=(745)/(760)=0.98atm`

conversion used : (1 atm = 760 torr)

T = temperature of the gas =
`20^oC=273+20=293K`

R = gas constant = 0.0821 Latm/moleK

V = volume of gas = ?

Now put all the given values in the above equation, we get :

`(0.98atm)* V=(0.833mole)* (0.0821Latm/moleK)* (293K)`

`V=0.2690L`

Therefore, the volume of
`H_2` will be, 0.2690 L

Answer 2

V = 20.4 L

Step-by-step explanation:

Step 1: Write the balanced equation

4 H₂O(g) + 3 Fe(s) ⟶ Fe₃O₄(s) + 4 H₂(g)

Step 2: Find the moles of H₂

We can establish the following relations.

• The molar mass of H₂O is 18.0 g/mol
• The molar ratio of H₂O to H₂ is 4:4.

The moles of H₂ obtained from 15.0 g of H₂O is:

`15.0gH_(2)O.(1molH_(2)O)/(18.0gH_(2)O) .(4molH_(2))/(4molH_(2)O) =0.833molH_(2)`

Step 3: Find the volume of H₂

We will use the ideal gas equation.

P = 745 torr × (1 atm/760 torr) = 0.980 atm

V = ?

n = 0.833 mol

R = 0.08206 atm.L/mol.K

T = 20°C + 273 = 293 K

P × V = n × R × T

0.980 atm × V = 0.833 mol × (0.08206 atm.L/mol.K) × 293 K

V = 20.4 L