Question

Suppose of iron(II) bromide is dissolved in of a aqueous solution of silver nitrate. Calculate the final molarity of bromide anion in the solution. You can assume the volume of the solution doesn't change when the iron(II) bromide is dissolved in it. Round your answer to significant digits.

Answer 1

The question is incomplete, here is the complete question:

Suppose 3.00 g of iron(II) bromide is dissolved in 350 mL of a 0.042 M aqueous solution of silver nitrate. Calculate the final molarity of bromide anion in the solution. You can assume the volume of the solution doesn't change when the iron(II) bromide is dissolved in it. Round your answer to 2 significant digits.

Answer: The final molarity of bromide ions in the solution is 0.042 M

Explanation:

• For iron (II) bromide:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Given mass of iron (II) bromide = 3 g

Molar mass of iron (II) bromide = 215.65 g/mol

Putting values in above equation, we get:

$\text{Moles of iron (II) bromide}=\frac{3g}{215.65g/mol}=0.014mol$

• For silver nitrate:

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$     .....(1)

Molarity of silver nitrate solution = 0.042 M

Volume of solution = 350 mL = 0.350 L     (Conversion factor:  1 L = 1000 mL)

Putting values in equation 1, we get:

$0.042M=\frac{\text{Moles of silver nitrate}}{0.350L}\\\\\text{Moles of silver nitrate}=(0.042mol/L\times 0.350L)=0.0147mol$

The chemical equation for the reaction of iron (II) bromide and silver nitrate follows:

$FeBr_2+2AgNO_3\rightarrow Fe(NO_3)_2+2AgBr$

By Stoichiometry of the reaction:

2 moles of silver nitrate reacts with 1 mole of iron (II) bromide

So, 0.0147 moles of silver nitrate will react with = $\frac{1}{2}\times 0.0147=0.00735mol$ of iron (II) bromide

As, given amount of iron (II) bromide is more than the required amount. So, it is considered as an excess reagent.

Thus, silver nitrate is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

2 moles of silver nitrate produces 2 moles of silver bromide

So, 0.0147 moles of silver nitrate will produce = $\frac{2}{2}\times 0.0147=0.0147moles$ of silver bromide

1 mole of silver bromide produces 1 mole of silver ions and 1 mole of bromide ions

Now, calculating the molarity of bromide ions by using equation 1:

Moles of bromide ions = 0.0147 moles

Volume of solution = 0.350 L

Putting values in equation 1, we get:

$\text{Molarity of }Br^-\text{ ions}=\frac{0.0147mol}{0.350L}\\\\\text{Molarity of }Br^-\text{ ions}=0.042M$

Hence, the final molarity of bromide ions in the solution is 0.042 M