g You are given a 1.25 gram mixture of calcium nitrate and calcium chloride. You dissolve this mixture in 200.0 mL of water and add an excess of 0.300 M silver nitrate. You collect and dry the white precipitate which forms and find it has a mass of 0.535 grams. Calculate the percent calcium chloride by mass in the original mixture.

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Answer:

16.51%

Explanation:

The reaction that takes place is

  • CaCl₂(aq) + 2AgNO₃(aq) → 2AgCl(s) + Ca(NO₃)₂(aq)

Meaning that the white precipitate that formed is AgCl. Now we convert 0.535 g of AgCl into moles, using its molar mass:

  • 0.535 g ÷ 143.43 g/mol = 0.00373 mol AgCl

Then we convert 0.00373 moles of AgCl into moles of CaCl₂, using the stoichiometric coefficients:

  • 0.00373 mol AgCl * [tex]\frac{1molCaCl_2}{2molAgCl}[/tex] = 0.00186 mol CaCl₂

We convert moles of CaCl₂ into grams, using its molar mass:

  • 0.00186 mol CaCl₂ * 110.98 g/mol = 0.206 g

Finally we calculate the percent of CaCl₂ by mass in the original mixture:

  • 0.206 g / 1.25 g * 100% = 16.51%

The percent calcium chloride by mass in the original mixture is 16.4%

The equation of the reaction between calcium nitrate and silver chloride is:

[tex]\mathbf{CaCl_2 + 2AgNO_3 \to 2 AgCl+Ca(NO_3)_2}[/tex]

Given that;

  • the weight mass of the white precipitate formed is = 0.535 grams

the number of moles of the precipitated AgCl is:

= 0.535 g / 143.32 g/mol

= 0.0037 moles

From the above reaction, If 2 moles of AgCl are formed  by 1 mole of CaCl2

Then, 0.0037 moles of AgCl will form (0.0037 × 1)/2 moles of CaCl2.

0.0037 moles of AgCl will form 0.00185 moles of CaCl2.

Now, we can say that the number of moles of CaCl2 present in the mixture is =  0.00185 moles

Mass amount of CaCl2 present = 0.00185 moles × 110.98 g/mol

Mass amount of CaCl2 present = 0.205 grams

Finally, the mass percentage   [tex]\mathbf{=\dfrac{0.205}{1.25}\times 100\%}[/tex]

= 16.4%

Therefore, we can conclude that the percent calcium chloride by mass in the original mixture is 16.4%

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