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A 0.7011-g mixture of sodium chloride and sodium nitrate is completely dissolved in water. When silver nitrate is added to the solution, 0.9805 g of solid precipitate. What is the percent by mass of chloride in the initial mixture?

Respuesta :

maacastrobr

Answer:

34.59 %

Explanation:

Sodium chloride dissolves in water as follows:

  • NaCl(s) → Na⁺(aq) + Cl⁻(aq)

Chloride ions are then precipitated with silver ions:

  • Ag⁺(aq) + Cl⁻(aq) → AgCl(s)

Now we use the give mass of AgCl to calculate the moles of Cl:

  • 0.9805 g AgCl ÷ 143.32 g/mol = 6.841x10⁻³ mol AgCl

mol AgCl = mol Cl

  • 6.841x10⁻³ mol Cl * 35.45 g/mol = 0.2425 g Cl

Finally we calculate the mass percent of Cl in the initial mixture:

  • 0.2425g / 0.7011g *100% = 34.59 %

The percent mass of chlorine in the initial mixture of 0.7011-g of sodium chloride and sodium nitrate is 34.59 %.

How do we calculate mass percent?

Mass percent of any substance in solution will be calculated as:

% mass = (mass of solute/mass of solution)×100%

To calculate the % mass of chlorine first we have to determine the initial mass of chlorine atom. Formation of precipitate after adding silver nitrate to the mixture of NaCl & NaNO₃ will be done according to the below equation:

Ag⁺(aq) + Cl⁻(aq) → AgCl(s)

From the stoichiometry of the above reaction, it is clear that same moles of chlorine and AgCl is involved in the reaction.

Moles of AgCl will be calculated as:

n = W/M, where

W = given mass = 0.9805 g

M = molar mass = 143.32 g/mol

n = 0.9805g / 143.32g/mol = 6.841x10⁻³ mol

So, initial moles of chlorine = 6.841x10⁻³ mol

Mass of chlorine = (6.841x10⁻³mol)(35.45g/mol) = 0.2425 g

Now % mass of chlorine will be calculated as:
% mass = (0.2425g / 0.7011g)100% = 34.59 %

Hence percent mass of chlorine is 34.59 %.

To know more about % mass, visit the below link:
https://brainly.com/question/13896694

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