Answer :
Answer
The mass in grams of AgBr formed = 3.84 g
Explanation
The given parameters are:
Volume of AgNO3 = 50.5 mL
Molarity of AgNO3 = 0.405 M
What to find:
The mass in grams of AgBr formed.
Step-by-step solution:
Step 1: Write a balanced equation for the reaction.
The balanced chemical equation for AgNO3 treated with an excess of aqueous hydrobromic acid is
[tex]AgNO_3(aq)+HBr(aq)\rightarrow AgBr(s)+HNO_3(aq)[/tex]Step 2: Convert the volume of 0.405 M AgNO3 that reacted to moles.
The moles of AgNO3 that reacted can be determined using the molarity formula.
[tex]Molatrity=\frac{Moles}{Volume\text{ }in\text{ }L}[/tex]The volume of AgNO3 needs to be converted from mL to L using the conversion formula below.
Conversion factor: 1000 mL = 1 L
50.5 mL = (50.5 mL/1000 mL) x 1 L = 0.0505 L
So,
[tex]\begin{gathered} 0.405M=\frac{Moles}{0.0505L} \\ \\ Moles=0.405M\times0.0505L \\ \\ Moles=0.0204525\text{ }mol \end{gathered}[/tex]Step 3: calculated the moles of AgBr formed.
Using the mole ratio of AgNO3 and AgBr in step 1 and the moles of AgNO3 in step 2; the moles of AgBr is calculated as shown below.
[tex]\begin{gathered} 1mol\text{ }AgNO_3=1mol\text{ }AgBr \\ \\ 0.0204525mol\text{ }AgNO_3=x \\ \\ Cross\text{ }multiply\text{ }and\text{ }divide\text{ }both\text{ }sides\text{ }by\text{ }1mol\text{ }AgNO_3 \\ \\ x=\frac{0.0204525mol\text{ }AgNO_3}{1mol\text{ }AgNO_3}\times1mol\text{ }AgBr \\ \\ x=0.0204525\text{ }mol\text{ }AgBr \end{gathered}[/tex]Step 4: Convert the moles of AgBr formed in step 3 to mass in grams.
Note that the molar mass of AgBr = 187.77 g/mol
Mass in grams of AgBr formed can be calculated using the mole formula.
[tex]\begin{gathered} Moles=\frac{Mass}{Molar\text{ }mass} \\ \\ Mass=Moles\times Molar\text{ }mass \\ \\ Mass=0.0204525\text{ }mol\times187.77\text{ }g\text{/}mol \\ \\ Mass=3.84\text{ }g \end{gathered}[/tex]Therefore, the mass in grams of AgBr formed when 50.5 mL of 0.405 M AgNO3 is treated with an excess of aqueous hydrobromic acid is 3.84 g.
