Question

Cacl2+2AgNO3(aq)-> Ca(No3)2 (aq)+2AgCl (s) if 25.0mL of a 2.00 m CaCl2 solution is used for the reaction shown above how many miles of chloride ions were involved in the reaction? Moles

Answer 1

Answer:

$n_{Cl^-}=0.1molCl^-$

Explanation:

Hello,

In this case, given the 2.00 M solution, we can compute the moles of calcium chloride that reacted:

$n_{CaCl_2}=2.00\frac{mol}{L} *25.0mL*\frac{1L}{1000mL}=0.05molCaCl_2$

Then, since in one mole of calcium chloride, we find two moles of chloride ions (see subscript), we can compute the moles of chloride ions that were involved in the reaction as shown below:

$n_{Cl^-}=0.05molCaCl_2*\frac{2molCl^-}{1molCaCl_2}\\ \\n_{Cl^-}=0.1molCl^-$

Best regards.

Answer 2

Answer:

0.10 mole

Explanation:

The number of mole of chloride ions involved in the reaction would be 0.10.

First, the number of moles of $CaCl_2$ needs to be calculated:

Molarity of $CaCl_2$ used = 2.00 M

Volume of $CaCl_2$ used = 25.0 mL = 0.025 L

Number of moles of $CaCl_2$ = molarity x volume

                               = 0.025 x 2 = 0.05 mole

$CaCl_2$ ionizes in aqueous solutions according to the following equation:

$CaCl_2 (aq) --> Ca^{2+} (aq) + 2Cl^- (aq)$

1 mole of $CaCl_2$ ionizes to give 2 moles of chloride ions.

There, 0.05 mole of $CaCl_2$ will produce:

      0.05 x 2 = 0.10 mole of $CaCl_2$