moles of Al=4.40 g/26.9815 g/mol=0.163
moles cl2=15.4g/70.906g/mol=0.217
the ratio is 2:3
cl2 is the limiting reagent
2Al + 3Cl2 ---> 2AlCl3
In the chemical reaction 2AlCl3 equals 2Al plus 3Cl2, the substances Al and Cl2 are called products. They are formed as a result of the chemical reaction between aluminum chloride (AlCl3).
This is a redox reaction, specifically a double displacement reaction where aluminum bromide and chlorine gas react to form aluminum chloride and bromine gas. In this reaction, aluminum undergoes a change in oxidation state, from +3 to 0, while chlorine goes from 0 to -1.
To find the limiting reactant, calculate the moles of each reactant using their molar masses. The balanced chemical equation is 2Al + 3Cl2 → 2AlCl3, so the stoichiometry ratio is 2:3. Aluminum is the limiting reactant as it forms fewer moles of product. Thus, all 25.0 g of aluminum will react with 112.5 g of chlorine to form 67.5 g of aluminum chloride.
2Al + 3Cl2 -> 2AlCl3
The balanced chemical equation for this reaction is: 2AlBr3 + 3Cl2 → 2AlCl3 + 3Br2.
3
The chemical equation for this reaction is: 2 AlBr3 + 3 Cl2 -> 2 AlCl3 + 3 Br2 It is a redox reaction where aluminum bromide is oxidized to aluminum chloride and chlorine is reduced to bromine.
The balanced chemical equation for the reaction between aluminum and chlorine is 2Al + 3Cl2 -> 2AlCl3. To find the limiting reactant, calculate the moles of each reactant using their molar masses. The reactant that produces the least amount of product is the limiting reactant. In this case, determine that with 28.0g of aluminum and 33.0g of chlorine, aluminum is the limiting reactant. Use the stoichiometry of the balanced equation to find the maximum mass of aluminum chloride that can be formed.
The balanced chemical equation for this reaction is: 2AlBr3 + 3Cl2 -> 2AlCl3 + 3Br2.
The equation for the formation of aluminum chloride from its elements is: 2Al(s) + 3Cl2(g) -> 2AlCl3(s)
2Al(s) + 3Cl2(g) -> 2AlCl3(s)