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A 2.89 g sample of a metal chloride, MCl2, is dissolved in water and treated with excess aqueous silver nitrate. The silver chloride that formed weighed 6.41 g. Calculate the molar mass of M. Question 3 options: 70.9 g/mol 29 g/mol 58.0 g/mol 65 g/mol 72.4 g/mol

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Answer:

58.0 g/mol

Explanation:

The reaction that takes place is:

  • MClâ‚‚ + 2AgNO₃ → 2AgCl + M(NO₃)â‚‚

First we calculate how many moles of silver chloride were produced, using its molar mass:

  • 6.41 g AgCl ÷ 143.32 g/mol = 0.0447 mol AgCl

Then we convert AgCl moles into MClâ‚‚ moles, using the stoichiometric ratio:

  • 0.0447 mol AgCl * [tex]\frac{1molMCl_2}{2molAgCl}[/tex] = 0.0224 mol MClâ‚‚

Now we calculate the molar mass of MClâ‚‚, using the original mass of the sample:

  • 2.86 g / 0.0224 mol = 127.68 g/mol

We can write the molar mass of MClâ‚‚ as:

  • Molar Mass MClâ‚‚ = Molar Mass of M + (Molar Mass of Cl)*2
  • 127.68 g/mol = Molar Mass of M + (35.45 g/mol)*2

Finally we calculate the molar mass of M:

  • Molar Mass of M = 57 g/mol

The closest option is 58.0 g/mol.

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