In the reaction #Zn + 2HCl -> ZnCl_2 + H_2#, how many moles of hydrogen will be formed when 4 moles of #HCl# are consumed?

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Jan 14, 2016


#"2 moles H"_2#


Your tool of choice here will be the mole ratio that exists between zinc metal, #"Zn"#, and hydrochloric acid, #":HCl"#, in the balanced chemical equation.

#"Zn"_text((s]) + color(red)(2)"HCl"_text((aq]) -> "ZnCl"_text(2(aq]) + "H"_text(2(g]) uarr#

You're dealing with a single replacement reaction in which zinc displaces the hydrogen from hydrochloric acid. The products of the reaction are aqueous zinc chloride and hydrogen gas.

Now, as you can see from the balance chemical equation, a #1:color(red)(2)# mole ratio exists between the two reactants.

This tells you that in order for the reaction to take place, you need to have twice as many moles of hydrochloric acid as you do of zinc metal.

At the same time, you have a #color(red)(2):1# mole ratio between hydrochloric acid and hydrogen gas.

This means that the reaction will always produce half as many moles of hydrogen gas as you have moles of hydrochloric acid.

Since you know that #4# moles of hydrochloric acid are taking part in the reaction, and assuming that you have enough zinc metal so that it doesn't act as a limiting reagent, you can say that the reaction will produce

#4 color(red)(cancel(color(black)("moles HCl"))) * "1 mole H"_2/(color(red)(2)color(red)(cancel(color(black)("moles HCl")))) = color(green)("2 moles H"_2)#

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