How do you balance the equation Al + HCl ? AlCl3 + H2?

2Al(s)+6HCl(aq)→2AlCl3(aq)+3H2(g)

Steps:

  • There are 3 Cl atoms on the product side and 1 on the reactant side. Add a coefficient of 3 in front of HCl.

Al(s)+3HCl(aq)→AlCl3(aq)+H2(g)

  • There are 3 H atoms on the reactant side and 2 on the product side. The coefficient in front of HCl from 3 to 6, and add a coefficient of 3 in front of the H2.

Al(s)+6HCl(aq)→AlCl3(aq)+3H2(g)

  • Here are now 6 Cl on the reactant side and 3 on the product side. Add a coefficient of 2 in front of AlCl3.

Al(s)+6HCl(aq)→2AlCl3(aq)+3H2(g)

  • There is 1 Al on the reactant side and 2 Al on the product side. Add a coefficient in front of Al on the reactant side.

    2Al(s)+6HCl(aq)→2AlCl3(aq)+3H2(g)

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