Question

A 0.999-g sample of a metal chloride, mcl2, is dissolved in water and treated with excess aqueous silver nitrate. The silver chloride that formed weighed 1.286 g. Calculate the atomic mass of m.

Answer 1

Atomic mass of M is $\boxed{151.6\text{ g}}$ .

Further Explanation:

Given reaction occurs as follows:

$\text{MCl}_2(aq)+2\text{AgNO}_3(aq)\rightarrow\text{2AgCl}(s)+\text{M}\left(\text{NO}_3}\right)_2(aq)$

The formula to calculate moles of AgCl is as follows:

$\text{Moles of AgCl}=\dfrac{\text{Mass of AgCl}}{\text{Molar mass of AgCl}}$                ...... (1)

Substitute $1.286\text{ g}$ for mass of AgCl and $143.32\text{ g/mol}$ for molar mass of AgCl in equation (1).

$\begin{aligned}\text{Moles of AgCl}=&\dfrac{\text{1.286 g}}{\text{143.32 g/mol}}\\=&0.00897\text{ mol}\end{aligned}$

According to the balanced chemical reaction, one mole of $\text{MCl}_2$ reacts with two moles of $\text{AgNO}_3$ to form two moles of AgCl and one mole of $\text{M}\left(\text{NO}_3\right)_2$, So stoichiometric ratio between $\text{MCl}_2$  and AgCl is 1:2.

Since two moles of AgCl are produced by one mole of $\text{MCl}_2$ , number of moles of $\text{MCl}_2$ that can produce 0.00897 moles of AgCl can be calculated as follows:

$\begin{aligned}\text{Moles of MCl}_2=&\left(\dfrac{\text{1 mol of MCl}_2}{\text{2 mol of AgCl}}\right)\left(0.00897\text{mol AgCl}\right)\\=&0.00449\text{ mol}\end{aligned}$

The formula to calculate number of moles of $\text{MCl}_2$ is as follows:

$\text{Moles of MCl}_2=\dfrac{\text{Mass of MCl}_2}{\text{Molar mass of MCl}_2}$                              ...... (2)

Rearrange equation (2) to calculate molar mass of $\text{MCl}_2$ .${\text{Molar mass of MCl}_2}=\dfrac{\text{Mass of MCl}_2}{\text{Moles of MCl}_2}$                                ...... (3)

Substitute 0.999 g for mass of $\text{MCl}_2$ and 0.00449 mol for moles of $\text{MCl}_2$  in equation (3).

$\begin{aligned}{\text{Molar mass of MCl}_2}=&\dfrac{\text{0.999 g}}{\text{0.00449 mol}}\\=&222.5\text{ g/mol}\end{aligned}$

The formula to calculate molar mass of $\text{MCl}_2$  is as follows:

$\text{Molar mass of MCl}_2=\left[1\left(\text{Atomic mass of M}\right)+\\\\2\left(\text{Atomic mass of Cl}\right)\right]}$  ....... (4)

Rearrange equation (4) to calculate atomic mass of M.

$\text{Atomic mass of M}=\left[\text{Molar mass of MCl}_2-2\left(\text{Atomic mass of Cl}\right)\right]$   ...... (5)

Substitute 222.5 g/mol for molar mass of $\text{MCl}_2$ and 35.453 g for atomic mass of Cl in equation (5).

$\begin{aligned}\text{Atomic mass of M}=&222.5\text{ g}-2\left(35.453\text{ g}\right)\\=&151.6\text{ g}\end{aligned}$

Therefore atomic mass of M is 151.6 g.

Learn more:

1. Calculate the moles of chlorine in 8 moles of carbon tetrachloride: brainly.com/question/3064603

2. Calculate the moles of ions in the solution: brainly.com/question/5950133

Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Mole concept

Keywords: AgCl, MCl2, 151.6 g, atomic mass, molar mass, 222.5 g/mol, 35.453 g, 1.286 g, 0.999 g, stoichiometric ratio, 1:2.

Answer 2

Mass of silver chloride formed = 1.286 g

Molar mass of silver chloride = 143.32 g/mol AgCl

Converting mass to moles of AgCl using the molar mass:

$1.286 g AgCl *\frac{1mol}{143.32g}= 0.00897 mol AgCl$

The balanced chemical equation between metal chloride and silver nitrate can be represented as,

$MCl_{2}(aq)+2AgNO_{3}(aq)-->2AgCl(s)+M(NO_{3})_{2}(aq)$

Calculating the moles of MCl2 from moles of AgCl that form:

$0.00897 mol AgCl*\frac{1molMCl_{2} }{2molAgCl} =0.00449mol MCl_{2}$

The given mass of $MCl_{2}$=0.999 g

Molar mass of $MCl_{2}$=$\frac{0.999g}{0.00449mol} =222.5g/mol$

Molar mass of M= Molar mass of $MCl_{2}$- 2 (Molar mass of Cl)

                          =151.6g/mol