Answer : The volume of
will be, 0.2690 L
Solution :
(a) Steps involved for this problem are :
First we have to calculate the moles of
.
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
.
![\text{Moles of }H_2O=\frac{\text{Mass of }H_2O}{\text{Molar mass of }H_2O}=\frac{15g}{18g/mole}=0.833moles](https://tex.z-dn.net/?f=%5Ctext%7BMoles%20of%20%7DH_2O%3D%5Cfrac%7B%5Ctext%7BMass%20of%20%7DH_2O%7D%7B%5Ctext%7BMolar%20mass%20of%20%7DH_2O%7D%3D%5Cfrac%7B15g%7D%7B18g%2Fmole%7D%3D0.833moles)
The balanced chemical reaction is,
![4H_2O(g)+3Fe(s)\rightarrow Fe_3O_4(s)+4H_2(g)](https://tex.z-dn.net/?f=4H_2O%28g%29%2B3Fe%28s%29%5Crightarrow%20Fe_3O_4%28s%29%2B4H_2%28g%29)
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](https://tex.z-dn.net/?f=PV%3DnRT)
where,
n = number of moles of gas = 0.833 mole
P = pressure of the gas = ![745torr=\frac{745}{760}=0.98atm](https://tex.z-dn.net/?f=745torr%3D%5Cfrac%7B745%7D%7B760%7D%3D0.98atm)
conversion used : (1 atm = 760 torr)
T = temperature of the gas = ![20^oC=273+20=293K](https://tex.z-dn.net/?f=20%5EoC%3D273%2B20%3D293K)
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)](https://tex.z-dn.net/?f=%280.98atm%29%5Ctimes%20V%3D%280.833mole%29%5Ctimes%20%280.0821Latm%2FmoleK%29%5Ctimes%20%28293K%29)
![V=0.2690L](https://tex.z-dn.net/?f=V%3D0.2690L)
Therefore, the volume of
will be, 0.2690 L