Question

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

Answer 1

Answer:

58.0 g/mol

Explanation:

The reaction that takes place is:

• MCl₂ + 2AgNO₃ → 2AgCl + M(NO₃)₂

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 MCl₂ moles, using the stoichiometric ratio:

• 0.0447 mol AgCl * $\frac{1molMCl_2}{2molAgCl}$ = 0.0224 mol MCl₂

Now we calculate the molar mass of MCl₂, using the original mass of the sample:

• 2.86 g / 0.0224 mol = 127.68 g/mol

We can write the molar mass of MCl₂ as:

• Molar Mass MCl₂ = 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.