Question

How many grams of AgBr would have been formed from 0.100 g of${\mathbf{RaBr}}_{\mathbf{2}}$?

Short answer

When silver nitrate combines with radium bromide, the quantity of silver bromide generated is 0.0972g

Step-by-step solution

Concept used

The volume of solution in litres divided by the amount of solute in moles: M stands for moles of solute per litre of solution.

The formula for calculating a solution's molarity is:

$\mathbf{Molarity}\mathbf{}\mathbf{\left(}\mathbf{M}\mathbf{\right)}\mathbf{=}\frac{\mathbf{Moles}\mathbf{}\mathbf{of}\mathbf{}\mathbf{solute}\mathbf{\left(}\mathbf{inmol}\mathbf{\right)}}{\mathbf{Volume}\mathbf{}\mathbf{ofsolution}\mathbf{(}\mathbf{in}\mathbf{}\mathbf{L}\mathbf{)}}$

Determine the amount of  AgBr

When silver nitrate interacts with radium bromide, the quantity of silver bromide generated is calculated.

Consider ${\mathrm{RaBr}}_2\mathrm{of}0.100\mathrm{g}$

The equation for the reaction of silver nitrate with radium bromide is

$$\begin{gathered} {\mathrm{RaBr}}_{2\left(\mathrm{aq}\right)}+2{\mathrm{AgNO}}_{3\left(\mathrm{aq}\right)}\to 2{\mathrm{AgBr}}_{\left(\mathrm{s}\right)}+\mathrm{Ra}{\left({\mathrm{NO}}_3\right)}_{2\left(\mathrm{aq}\right)} \\ \frac{0.100\mathrm{g}}{386\mathrm{g}/\mathrm{mol}}=2.591\times {10}^{-4}\mathrm{mol} \\ =2\times 2.591\times {10}^{-4}\mathrm{mol} \end{gathered}$$

Two moles of silver bromide are produced from one mole of radium bromide.

The number of silver bromide moles

$$\begin{gathered} 2\times 2.591\times {10}^{-4}\mathrm{mol}=5.181\times {10}^{-4}\mathrm{mol} \\ \mathrm{Molarmass}=\frac{\mathrm{Mass}}{\mathrm{Mole}} \\ \mathrm{Mass}=187.77\mathrm{g}/\mathrm{mol}\times 5.181\times {10}^{-4}\mathrm{mol} \\ =0.0972\mathrm{g} \end{gathered}$$

Therefore, silver nitrate combines with radium bromide and the quantity of silver bromide formed is 0.0972 g