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:

V = 20.4 L

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:

$15.0gH_{2}O.\frac{1molH_{2}O}{18.0gH_{2}O} .\frac{4molH_{2}}{4molH_{2}O} =0.833molH_{2}$

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

Answer 2

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 $H_2O$.

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

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 = $745torr=\frac{745}{760}=0.98atm$

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 :

$(0.98atm)\times V=(0.833mole)\times (0.0821Latm/moleK)\times (293K)$

$V=0.2690L$

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