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Zinc and hydrochloric acid react according to the reaction: `Zn_((s))+2HCl_((aq.))rarr ZnCl_(2(aq.))+H_(2(g))` If `0.30` mole of `Zn` are added to hydrochloric acid containing `0.52` mole `HCl`, how many moles of `H_(2)` are produced? |
Answer» The balanced chemical equation for the reaction is : `underset("1 mol")(Zn(s))+underset("2 mol")(2HCl(aq))rarrZnCl_(2)(aq)+underset("1 mol")(H_(2)(g))` Step I. Determine of limiting reactant 1 mole of Zn react with HCl = 2 mol 0.30 mole of Zn react with HCl `= 2 xx 0.3 = 0.6 mol` But HCl actually present in solution `= 0.52` mol This means that HCl is present in lesser amount than what is actually required. `:.` It is the limiting reactant Step II. Determination of moles hydrogen produced 2.0 mole of HCl on reacting with Zn produce `H_(2)(g)=1` mol 0.52 mole of HCl on reacting with Zn produce `H_(2)(g)=(1xx 0.52)/(2)=0.26` mol. |
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