Knowing that 2 aluminum atoms and 3 chlorine molecules react to produce 2 units of aluminum chloride, how many moles of aluminum chloride can be produced by reacting 7.8 moles of chlorine with excess aluminum?
I think it is clearer if you write an equation. 2Al + 3Cl2 ==> 2AlCl3
Now use the coefficients to convert moles Cl2 to moles AlCl3.
To determine the number of moles of aluminum chloride produced, we need to use the balanced chemical equation for the reaction.
The given balanced chemical equation is:
2 Al + 3 Cl2 → 2 AlCl3
According to the equation, 2 moles of aluminum react with 3 moles of chlorine to produce 2 moles of aluminum chloride.
We are given that there are 7.8 moles of chlorine available for the reaction. However, we need to figure out how many moles of aluminum chloride can be produced by this quantity of chlorine.
To calculate the number of moles of aluminum chloride produced, we can set up a ratio:
(3 moles of Cl2 / 2 moles of AlCl3) = (7.8 moles of Cl2 / x)
This ratio represents the stoichiometry of the reaction.
Now, we can cross multiply the ratio and solve for x:
3 moles of Cl2 * x = 2 moles of AlCl3 * 7.8 moles of Cl2
3x = 15.6
Dividing both sides by 3, we find:
x = 5.2 moles of AlCl3
Therefore, 5.2 moles of aluminum chloride can be produced by reacting 7.8 moles of chlorine with excess aluminum.