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This equation is Al2O3 + 6 HCl = 2 AlCl3 + 3 H2O.
2 AlCl3 -> 2 Al + 3 Cl2
3Ag2SO4 + 2AlCl3 -> 6AgCl + Al2(SO4)3
2C2H6 + 7O2 -> 4CO2 + 6H2O By looking at the ethane I could see that I needed to put the 2 coefficient in front of it to make the oxygens come out even. Always balance O2 last in these combustion reactions.
3
The balanced equation is Al(OH)3 + 3HCl ==> 3H2O + AlCl3
Al(OH)3 + 3HCl >> AlCl3 + 3H2O
The balanced equation looks like this: k2so4+alcl3 kcl+al2(5o4)3
This equation is Al2O3 + 6 HCl = 2 AlCl3 + 3 H2O.
2 AlCl3 -> 2 Al + 3 Cl2
3Ag2SO4 + 2AlCl3 -> 6AgCl + Al2(SO4)3
Al2O3 + 6 HCl --> 2 AlCl3 + 3 H2
2C2H6 + 7O2 -> 4CO2 + 6H2O By looking at the ethane I could see that I needed to put the 2 coefficient in front of it to make the oxygens come out even. Always balance O2 last in these combustion reactions.
4 LiH + AlCl3 =======> LiAlH4 + 3 LiCl
3
2 Al + 6 HCl = 2 AlCl3 + 3 H2
Balanced equation first. ( you have been told the limiting reactant. ) Al(OH)3 + 3HCl >> AlCl3 + 3H2O Molar mass Al(OH)3 = 78.004 grams/// Molar mass AlCl3 = 133.33 grams Conversion. 328g Al(OH)3 (1mol Al(OH)3/78.004g)(1 mol AlCl3/1mol Al(OH)3)(133.33g AlCl3/1mol AlCl3) =560.64 grams produced