Answer: 1. 
2. 3 moles of 
: 2 moles of 
3. 0.33 moles of : 0.92 moles of
4. is the limiting reagent and is the excess reagent.
5. Theoretical yield of 
is 29.3 g
Explanation:
To calculate the moles :
The balanced chemical equation is:
According to stoichiometry :
3 moles of require = 2 moles of
Thus 0.33 moles of will require=
of
Thus is the limiting reagent as it limits the formation of product and is the excess reagent.
As 3 moles of give = 2 moles of
Thus 0.33 moles of give = of
Theoretical yield of 
Thus 29.3 g of aluminium chloride is formed.
Could you show the characteristics?
Answer:
The <u>equilibrium constant</u> is:
Explanation:
The correct equation is:
Thus, with the equilibrium concentrations you can calculate the equilibrium constant, Kc.
The equation for the equilibrium constant is:
![k_c=\dfrac{[NH_3]^2}{[N_2]\cdot [H_2]^3}](https://tex.z-dn.net/?f=k_c%3D%5Cdfrac%7B%5BNH_3%5D%5E2%7D%7B%5BN_2%5D%5Ccdot%20%5BH_2%5D%5E3%7D)
Substituting:
Answer:
The half-life time, the team equired for a quantity to reduce to half of its initial value, is 79.67 seconds.
Explanation:
The half-life time = the time required for a quantity to reduce to half of its initial value. Half of it's value = 50%.
To calculate the half-life time we use the following equation:
[At]=[Ai]*e^(-kt)
with [At] = Concentration at time t
with [Ai] = initial concentration
with k = rate constant
with t = time
We want to know the half-life time = the time needed to have 50% of it's initial value
50 = 100 *e^(-8.7 *10^-3 s^- * t)
50/100 = e^(-8.7 *10^-3 s^-1 * t)
ln (0.5) = 8.7 *10^-3 s^-1 *t
t= ln (0.5) / -8.7 *10^-3 = 79.67 seconds
The half-life time, the team equired for a quantity to reduce to half of its initial value, is 79.67 seconds.
