X, Y and Z are all gases that behave ideally and react according to the equation shown.
X(g) + 2Y(g) 2Z(g)
When 3.0 mol of X and 3.0 mol of Y are placed inside a container with a volume of 1.0 dm3, they
react to form the maximum amount of Z.
The final temperature of the reaction vessel is 120 C.
What is the final pressure inside the reaction vessel?
Answers
Answer:
Moles of X = 3.0 moles
Moles of Y = 3.0 moles
For the given chemical equation:
X(g)+2Y(g)\rightarrow 2Z(g)X(g)+2Y(g)→2Z(g)
By Stoichiometry of the reaction:
2 moles of Y reacts with 1 mole of X
So, 3.0 moles of Y will react with = \frac{1}{2}\times 3.0=1.5mol
2
1
×3.0=1.5mol of X
As, given amount of 'X' is more than the required amount. So, it is considered as an excess reagent.
Thus, 'Y' is considered as a limiting reagent because it limits the formation of product.
By Stoichiometry of the reaction:
2 moles of Y produces 2 moles of Z
So, 3.0 moles of Y will produce = \frac{2}{2}\times 3.0=3.0mol
2
2
×3.0=3.0mol of X
Moles of X left in the vessel = 3.0 - 1.5 = 1.5 moles
To calculate the pressure, we use the equation given by ideal gas equation:
PV=nRTPV=nRT
where,
P = pressure of the gas
V = Volume of gas = 1.0dm^31.0dm
3
n = number of moles = Moles of Z + Moles of X left = [3.0 + 1.5] = 4.5 moles
R = Gas constant = 8.31dm^3\text{ kPa }mol^{-1}K^{-1}8.31dm
3
kPa mol
−1
K
−1
T = temperature of the gas = 102^oC=[102+273]=393K102
o
C=[102+273]=393K
Putting values in above equation, we get:
\begin{gathered}P\times 1.0dm^3=4.5mol\times 8.31dm^3\text{ kPa }mol^{-1}K^{-1}\times 393K\\\\P=1.47\times 10^4kPa\end{gathered}
P×1.0dm
3
=4.5mol×8.31dm
3
kPa mol
−1
K
−1
×393K
P=1.47×10
4
kPa
Converting the pressure in pascals, we use the conversion factor:
1kPa=1000Pa=10^3Pa1kPa=1000Pa=10
3
Pa
Converting into pascals, we get:
\begin{gathered}\Rightarrow 1.47\times 10^4kPa\times \frac{10^3Pa}{1kPa}\\\\\Rightarrow 1.47\times 10^7Pa\end{gathered}
⇒1.47×10
4
kPa×
1kPa
10
3
Pa
⇒1.47×10
7
Pa