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).
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 formula of aluminium chloride is AlCl3. The atomic weight of aluminium is 27 and that of chlorine is 35.5. That means 35.5*3 grams of chlorine will combine with 27 grams of aluminium. So 33 grams of chlorine will combine with 8.37 grams of aluminium. The addition of both makes it 41.37 grams. In this reaction, the whole chlorine will be utilized and only part of the aluminium.
The equation for the formation of aluminum chloride from its elements is: 2Al(s) + 3Cl2(g) -> 2AlCl3(s)
2Al + 3Cl2 -> 2AlCl3 aluminium reacts with chlorine gas to form aluminium trichloride.
2Al(s) + 3Cl2(g) -> 2AlCl3(s)
HCl + Al2O3 ----- H2O + AlCl3 H 1 Cl 1 Al 2 O 3 6HCl + Al2O3 --------- 3H2O + 2AlCl3