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).
The balanced chemical equation for this reaction is: 2AlBr3 + 3Cl2 -> 2AlCl3 + 3Br2.
2Al + 3Cl2 -> 2AlCl3
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.
The balanced chemical equation for this reaction is: 2AlBr3 + 3Cl2 → 2AlCl3 + 3Br2.
3
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 equation for the formation of aluminum chloride from its elements is: 2Al(s) + 3Cl2(g) -> 2AlCl3(s)
The balanced chemical equation for the reaction between aluminum metal and chlorine gas to form solid aluminum chloride is: 2Al (s) + 3Cl2 (g) -> 2AlCl3 (s)
2Al(s) + 3Cl2(g) -> 2AlCl3(s)
The balanced chemical equation for aluminum metal burning in chlorine gas to produce aluminum chloride is: 2Al(s) + 3Cl2(g) -> 2AlCl3(s)