Question

A mass of 0.6539 g of zinc is dissolved in excess hydrochloric acid, according to the following equation:

[ Zn_{(aq)} + 2HCl_{(aq)} → ZnCl_{2(aq)} + H_2_{(g)} ]
How many moles of (Cl^-) ions are formed in the reaction?
A: 0.1 moles
B: 0.01 moles
C: 0.002 moles
D: 0.02 moles

Answer 1

To determine the number of moles of Cl- ions formed when 0.6539 g of zinc reacts with excess hydrochloric acid, first calculate the moles of zinc, and then use stoichiometry from the balanced reaction equation. The result is 0.02 moles of Cl- ions.

Step-by-step explanation:

The question is asking how many moles of chloride ions (Cl-) are formed when a mass of 0.6539 g of zinc reacts with excess hydrochloric acid. The balanced equation for the reaction is:

Zn (s) + 2HCl(aq) → ZnCl2 (aq) + H2 (g)

To find the number of moles of Cl- ions formed, we need to calculate the moles of Zn that reacted and then use the stoichiometry of the reaction to find the moles of Cl-. Zinc has a molar mass of approximately 65.38 g/mol. Thus, the moles of zinc that reacted are:

0.6539 g Zn × (1 mol Zn / 65.38 g Zn) = 0.0100 moles Zn
For every mole of Zn that reacts, there are two moles of Cl- ions produced, according to the stoichiometry of the balanced equation. Therefore, the moles of Cl- ions are:

0.0100 moles Zn × (2 moles Cl- / 1 mole Zn) = 0.0200 moles Cl-

So the correct answer is D: 0.02 moles of Cl- ions are formed.