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An impure sample of table salt that weighed 0.8421 g, when dissolved in water and treated with excess AgNO3, formed 2.044 g of AgCl. what is the percentage of NaCl in the impure sample?

Respuesta :

superman1987

Answer:

99.24%.

Explanation:

  • NaCl reacted with AgNO₃ as in the balanced equation:

NaCl + AgNO₃ → AgCl(↓) + NaNO₃,

1.0 mol of NaCl reacts with 1.0 mol of AgNO₃ to produce 1.0 mol of AgCl and 1.0 mol of NaNO₃.

  • We need to calculate the no. of moles of AgCl produced:

no. of moles of AgCl = mass/molar mass = (2.044 g)/(143.32 g/mol) = 0.0143 mol.

  • Now, we can calculate the no. of moles of NaCl that can precipitated as AgCl (0.0143 mol), these moles represents the no. of moles of pure NaCl in the sample:

using cross multiplication:

1.0 mol of NaCl produce → 1.0 mol of AgCl, from the stichiometry.

∴ 0.0143 mol of NaCl produce → 0.0143 mol of AgCl.

  • Now, we can get the mass of puree NaCl in the sample:

mass of pure NaCl = (no. of moles of pure NaCl)(molar mass of NaCl) = (0.0143 mol)(58.44 g/mol) = 0.8357 g.

∴ The percentage of NaCl in the impure sample = [(mass of pure NaCl)/(mass of the impure sample)] x 100 = [(0.8357 g)/(0.8421 g)] x 100 = 99.24%.

dsdrajlin

An impure sample of table salt that weighed 0.8421 g and treated with excess AgNO₃ formed 2.044 g of AgCl, has a percentage of NaCl of 98.98%.

Let's consider the reaction between NaCl and AgNO₃ to produce AgCl and NaNO₃.

NaCl + AgNO₃ ⇒ AgCl + NaNO₃

We can calculate the mass of NaCl that produced 2.044 g of AgCl using the following relations.

  • The molar mass of AgCl is 143.32 g/mol.
  • The molar ratio of AgCl to NaCl is 1:1.
  • The molar mass of NaCl is 58.44 g/mol.

[tex]2.044 g AgCl \times \frac{1molAgCl}{143.32 g AgCl} \times \frac{1molNaCl}{1molAgCl} \times \frac{58.44gNaCl}{1molNaCl} = 0.8335gNaCl[/tex]

An impure sample of mass 0.8421 g contains 0.8335 g of NaCl. The percentage of NaCl in the impure sample is:

[tex]\% NaCl = \frac{0.8335g}{0.8421g} \times 100\% = 98.98\%[/tex]

An impure sample of table salt that weighed 0.8421 g and treated with excess AgNO₃ formed 2.044 g of AgCl, has a percentage of NaCl of 98.98%.

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