Given the balanced equation
2Al + 6HBr --> 2AlBr3 + 3H2
In order to find how many grams of HBr are required to produce 150g AlBr3, we must convert from mass to mass (mass --> mass conversion).
150g AlBr3 * 1 mol AlBr3 * 6 molecules HBr = 136.52 or 137g HBr
----------- 266.6g AlBr3 * 2 molecules AlBr3
AlBr3 is the product thus: 2 Al + 3 Br2 ------> 2 AlBr3 Aluminum has the anion charge of 3+ and a bromine atom has just 1+ thus you need 3 bromine atoms in order to make the product and end up with a non-charged molecule. Unfortunately bromine comes in its diatomic (two of the same atom) in this equation thus you can't have 1.5 of those molecules. If you multiply in such a way, in this case by two, to make everything whole numbers you have your equation.
2Fe+3Cl2------>2FeCl3
Al2(SO4)3 + 6KOH --> 2 [Al(OH)3](s) + 3K2SO4or in excess hydroxide:Al2(SO4)3 + 8KOH --> 2K+ + 2 [Al(OH)4]-(aq) + 3K2SO4
The equation is, 2Al + 3Br2 = 2AlBr3
2AlBr3 + 3Cl2 -> 2AlCl3 + 3Br2
Aluminium metal reacts with bromine gas to form aluminium tribromide. 2Al + 3Br2 ==> 2AlBr3
1313
Single Replacement
2AlBr3+3Cl2 --> 2AlCl3+3Br2
Al2O3 + 6HBr = 2Albr3 + 3H2O
2AlBr3 + 3Cl2 -> 2AlCl3 + 3Br2
2Al+3Br2= 2AlBr3 Aluminium reacts with bromine gas to form aluminium tribromide.
it gives us a new substance that is aluminum bromide therefore it is a chemical change.
AlBr3 is the product thus: 2 Al + 3 Br2 ------> 2 AlBr3 Aluminum has the anion charge of 3+ and a bromine atom has just 1+ thus you need 3 bromine atoms in order to make the product and end up with a non-charged molecule. Unfortunately bromine comes in its diatomic (two of the same atom) in this equation thus you can't have 1.5 of those molecules. If you multiply in such a way, in this case by two, to make everything whole numbers you have your equation.
First we need to determine whether this is a combination, combustion, or decomposition reaction.It's classified as a combination.Next we need to determine the correct formulas for the reactants.These would be Al(s) and Br2(l)Next, we need to identify the product of this reaction, which would be AlBr3Lastly, we need to balance the equation.Your final answer will be:2Al(s) + 3Br2(l) --> 2AlBr3(s)