Question

Zinc Oxide reacts with Hydrochloric acid to product zinc chloride and dihydrogen monoxide. If 3.6 moles of Zinc oxide reacts with an excess of Hydrochloric acid, then how many grams of Zinc Chloride were produced?

Answer 1

Hint: To solve this question, we’ll have to know the balanced chemical equation of the reaction given. By knowing the no. of moles of reactant needed to give the product, we can determine how much of product will be obtained.

Complete answer:
The reaction given to us is Zinc Oxide reacts with Hydrochloric acid to product zinc chloride and dihydrogen monoxide. The chemical equation can be given as:
$ZnO + HCl \to ZnC{l_2} + {H_2}O$
In here the no. of molecules of Hydrogen and Chlorine are not balanced. We’ll have to balance HCl by two to balance the equation. The balanced equation is:
$Zn{O_{(aq)}} + 2HCl \to ZnC{l_2}_{(aq)} + {H_2}{O_{(l)}}$
We can infer from the above equation that 1 mol of ZnO is required to give 1 mole of Zinc Chloride. We are given 3.6 Moles of ZnO, hence $3.6mol{\text{ }}ZnO = 3.6mol{\text{ }}ZnC{l_2}$
To find the amount (in grams) from no. of moles we’ll use the formula: $Moles = \dfrac{{Mass}}{{M.{M_{ZnC{l_2}}}}}$
Molar Mass of $ZnC{l_2} = 65 + 2 \times 35.5 = 136g/mol$
Therefore, the amount of $ZnC{l_2}$ formed $ = moles \times M.{M_{ZnC{l_2}}}$
$Mass(g) = 3.6 \times 136 = 489.6g$
Therefore, 489.6g of $ZnC{l_2}$ is formed.

Note:
No specific formula is required for solving this question. Only conversion of moles in grams is required. Remember that while converting moles into grams the molar mass should be g/mol, to obtain the mass in grams.
In here we were given that hydrochloric acid is in excess, which means that Zinc oxide is less, then Zinc oxide will be considered as the limiting reagent. The limiting reagent always finishes up in a reaction. If excess was not given then we would have required 2 moles of hydrochloric acid for the completion of the reaction.