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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

Respuesta :

sebassandin

Answer:

[tex]n_{Cl^-}=0.1molCl^-[/tex]

Explanation:

Hello,

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

[tex]n_{CaCl_2}=2.00\frac{mol}{L} *25.0mL*\frac{1L}{1000mL}=0.05molCaCl_2[/tex]

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:

[tex]n_{Cl^-}=0.05molCaCl_2*\frac{2molCl^-}{1molCaCl_2}\\ \\n_{Cl^-}=0.1molCl^-[/tex]

Best regards.

Eduard22sly

The number of mole of chloride ion, Cl¯ involved in the reaction is 0.1 mole

We'll begin by calculating the number of mole of CaCl₂ in the solution. This can be obtained as follow:

Volume = 25 mL = 25 / 1000 = 0.025 L

Molarity = 2 M

Mole of CaCl₂ =?

Mole = Molarity x Volume

Mole of CaCl₂ = 2 × 0.025

Mole of CaCl₂ = 0.05 mole

  • Finally, we shall determine the number of chloride ion, Cl¯ in the solution. This can be obtained as follow:

CaCl₂(aq) —> Ca²⁺(aq) + 2Cl¯(aq)

From the balanced equation above,

1 mole of CaCl₂ contains 2 moles of Cl¯.

Therefore,

0.05 mole of CaCl₂ will contain = 2 × 0.05 = 0.1 mole of Cl¯.

Thus, the number of mole of chloride ion, Cl¯ involved in the reaction is 0.1 mole

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