Balance this equation_Al +_HCl --> _H₂ +_AlCl₃
Answers
Explanation:
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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Answer:
Your answer is the first option, friend.
