Question

If 25.0 mL of a 2.00 M CaCl2 solution is used for the reaction
shown above, how many moles of chloride ions were involved
in the reaction?__moles

Answer 1

0.1 moles of chloride ions were involved  in the reaction if 25.0 mL of a 2.00 M CaCl2 solution is used for the reaction.

Explanation:

Data given:

volume of Ca$Cl_{2}$ = 25 ml 0r 0.025

molarity of the calcium chloride solution = 2M

number of chloride ions =?

Balance chemical reaction:

Ca + 2$Cl_{2}$ ⇒$Ca_{} Cl_{2}$

number of moles in 25 ml is calculated as:

molarity = $\frac{number of moles}{volume}$

number of moles of calcium chloride  = molarity x volume

putting the values in the equation:

number of moles = 2 x 0.025

                            = 0.05 moles of calcium chloride

1 mole of  Ca$Cl_{2}$  decomposes as 1 calcium ion and 2 chloride ions

so 0.05 moles will have x moles of chloride ion

$\frac{2}{1}$ = $\frac{x}{0.05}$

x= 0.1

0.1 moles of chloride ions will be involved in the reaction.

Answer 2

0.1 moles of chloride ions were involved in the reaction.

Considering the concentration of 2.0M of $CaCl_{2}$ and the use of 25 $ml$ of this compound we have that:

number of moles = $\frac{2 moles}{1000ml} \times 0.025ml$ = $0.05 moles$

We have that the amount of moles of $CaCl_{2}$ is equal to 0.05 moles, so, according to the equation that forms this compound we can say that:

                                       $Ca + 2Cl$⇒$CaCl2$

So, 1 mole of $CaCl2$ decomposes as 1 calcium ion and 2 chloride ions, that is, the number of moles of chlotide is twice the number of moles of CaCl2:
 
$0.05 \times 2 = 0.1 moles$      

In this way, 0.1 moles of chloride ions were involved in the reaction.

Learn more about mole calculation in: brainly.com/question/15321053