2Al(s) + 6HCl(aq) = 2AlCl3(aq) + 3H2(g).
The molar ratios are 2:6 :: 2:3
This means that two moles of aluminium will react to completion with exactly six moles of hydrocchloric acid. At completion there will be two moles of aluminium chloride and three moles of hydrogen gas.
So if you had 8 moles of hydrochloric acid and only 2 moles of aluminium, then the aluminium would be the limiting reactant, because from the 8 moles hydrochloric acid you are using only six moles, leaving two moies on hydrochloric acid unreacted.
Conversely, if you had 3 moles of Al and 6 moles HCl , then the HCl would be the limiting reactant. Because the 6 moles of acid would only react with two moles of Al leaving one mole Al unreacted.
The balanced equation for the reaction between aluminum and hydrochloric acid is: 2Al + 6HCl -> 2AlCl3 + 3H2
2Al + 6HCl >> 2AlCl3 + 3H2
2Al + 6HCl --> 2AlCl3 + 3H2
The balanced equation for hydrochloric acid reacting with aluminum metal is: 2Al(s) + 6HCl(aq) → 2AlCl3(aq) + 3H2(g).
The reaction is 2Al+ 6HCl --> 2AlCl3 + 3H2
The chemical equation for this reaction is: 2Al(s) + 6HCl(aq) -> 2AlCl3(aq) + 3H2(g)
2Al(s) + 6HCl(aq) ----- 2AlCl3(aq) + 3H2(g).
Al2O3 + 6HCl >> 2AlCl3 + 3H2O
2AlBr3(aq) + 3Cl2(g) = 2AlCl3(aq) + 3Br2(l)
The reaction between aluminum (Al) and hydrochloric acid (HCl) produces aluminum chloride (AlCl3) and hydrogen gas (H2). The balanced chemical equation for this reaction is: 2Al + 6HCl → 2AlCl3 + 3H2
6HCl + Al2(CO3)3 ==> 2AlCl3 + 3CO2 + 3H2O
To determine the limiting reactant, we need to calculate the amount of each reactant in terms of the same unit (moles). Then, we compare the mole ratio of the reactants from the balanced chemical equation. In this case, convert 60.0 grams of aluminum to moles and 25 grams of hydrochloric acid to moles. Compare the moles of each reactant to determine the limiting reactant.