Respuesta :
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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