Answer:
163.2g
Explanation:
First let us generate a balanced equation for the reaction. This is shown below:
4Al + 3O2 —> 2Al2O3
From the question given, were were told that 3.2moles of aluminium was exposed to 2.7moles of oxygen. Judging by this, oxygen is excess.
From the equation,
4moles of Al produced 2moles of Al2O3.
Therefore, 3.2moles of Al will produce = (3.2x2)/4 = 1.6mol of Al2O3.
Now, let us covert 1.6mol of Al2O3 to obtain the theoretical yield. This is illustrated below:
Mole of Al2O3 = 1.6mole
Molar Mass of Al2O3 = (27x2) + (16x3) = 54 + 48 =102g/mol
Mass of Al2O3 =?
Number of mole = Mass /Molar Mass
Mass = number of mole x molar Mass
Mass of Al2O3 = 1.6 x 102 = 163.2g
Therefore the theoretical of Al2O3 is 163.2g
Answer:
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Higher concentrations of reactants = More collisions between molecules = More possible reactions between the molecules = Higher reaction rate
<h3>What is concentration?</h3>
A solution is made up of two components, solute and solvent. In chemistry, we define the concentration of solution as the amount of solute dissolved in the solution.
Increasing the concentration of reactants generally increases the rate of reaction because more of the reacting molecules or ions are present to form the reaction products. This is especially true when concentrations are low and few molecules or ions are reacting.
Hence, Higher concentrations of reactants = More collisions between molecules = More possible reactions between the molecules = Higher reaction rate.
Learn more about the concentration here:
brainly.com/question/10725862
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Answer:
Explanation:
In this problem we only have information of the equilibrium, so we need to find a expression of the free energy in function of the constant of equilireium (Keq):
Being Keq:
![K_{eq}=\frac{[fructose][Pi]}{[Fructose-1-P]}](https://tex.z-dn.net/?f=K_%7Beq%7D%3D%5Cfrac%7B%5Bfructose%5D%5BPi%5D%7D%7B%5BFructose-1-P%5D%7D)
Initial conditions:
![[Fructose-1-P]=0.2M](https://tex.z-dn.net/?f=%5BFructose-1-P%5D%3D0.2M)
![[Fructose]=0M](https://tex.z-dn.net/?f=%5BFructose%5D%3D0M)
![[Pi]=0M](https://tex.z-dn.net/?f=%5BPi%5D%3D0M)
Equilibrium conditions:
![[Fructose-1-P]=6.52*10^{-5}M](https://tex.z-dn.net/?f=%5BFructose-1-P%5D%3D6.52%2A10%5E%7B-5%7DM)
![[Fructose]=0.2M-6.52*10^{-5}M](https://tex.z-dn.net/?f=%5BFructose%5D%3D0.2M-6.52%2A10%5E%7B-5%7DM)
![[Pi]=0.2M-6.52*10^{-5}M](https://tex.z-dn.net/?f=%5BPi%5D%3D0.2M-6.52%2A10%5E%7B-5%7DM)
Free-energy for T=298K (standard):
