25 / 102 = x / 267
x = 267 (25/102) = 65.4 g
2AlBr3(aq) + 3Cl2(g) = 2AlCl3(aq) + 3Br2(l)
2Al + 3O2 --> 2Al2O3 0.78 moles O2 (2 moles Al2O3/3 moles O2) 0.52 moles Al2O3 produced ===========================( assuming oxygen limits )
The ionic compound Al2O2 forms when aluminum reacts with oxygen.
4Fe + 3O2 -> 2Fe2O3 is the balanced equation.It'll need enough oxygen (> (0.46*3)/4 mole O2)to give (0.46*2)/4 mole Fe2O3, so 0.23 mole Fe2O3is produced.
First you need a balanced equation: Al2O3 + 6HCl -> 2AlCl3 + 3H2O then you do a simple mole conversion: 1mol HCL/1 X 1mol Al2O3/ 6mol HCl which is (1 X 1) / (1 X 6) = 1/6mol Al2O3 is needed.
Al2O3 + 6HCl >> 2AlCl3 + 3H2O
2AlBr3(aq) + 3Cl2(g) = 2AlCl3(aq) + 3Br2(l)
HCl + Al2O3 ----- H2O + AlCl3 H 1 Cl 1 Al 2 O 3 6HCl + Al2O3 --------- 3H2O + 2AlCl3
al2o3
2Al + 3O2 --> 2Al2O3 0.78 moles O2 (2 moles Al2O3/3 moles O2) 0.52 moles Al2O3 produced ===========================( assuming oxygen limits )
The ionic compound Al2O2 forms when aluminum reacts with oxygen.
81.93%
4Fe + 3O2 -> 2Fe2O3 is the balanced equation.It'll need enough oxygen (> (0.46*3)/4 mole O2)to give (0.46*2)/4 mole Fe2O3, so 0.23 mole Fe2O3is produced.
First you need a balanced equation: Al2O3 + 6HCl -> 2AlCl3 + 3H2O then you do a simple mole conversion: 1mol HCL/1 X 1mol Al2O3/ 6mol HCl which is (1 X 1) / (1 X 6) = 1/6mol Al2O3 is needed.
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Fe2O3 + 2Al ===> Al2O3 + 2FeIn this reaction the number of moles of Al2O3 produced is dependent on the number of moles of Fe2O3 and Al that one starts with. For every 1 mole Fe2O3 and 2 moles Al, one gets 1 moles of Al2O3.
Yes, aluminum reacts with oxygen to form aluminum oxide. 2Al(s) + 3O2(g) ---------> Al2O3(s)