Answer:
[tex]Y=96.0\%[/tex]
Explanation:
Hello there!
In this case, considering the reaction between silver nitrate and sodium bromide, whereby sodium nitrate and silver bromide are produced:
[tex]NaBr+AgNO_3\rightarrow NaNO_3+AgBr[/tex]
By bearing to mind the solubility rules, we infer the precipitate is silver bromide, whose moles are computed by using the molarity and volume of each reactant and then decide the correct value according to the limiting reactant:
[tex]n_{AgBr}=0.0350L*0.100molAgNO_3*\frac{1molAgBr}{1molAgNO_3}=0.00350mol AgBr\\\\n_{AgBr}=0.0450L*0.080molNaBr*\frac{1molAgBr}{1molNaBr}=0.0036molAgBr[/tex]
Thus, since sodium nitrate produces the smallest theoretical yield, we calculate the grams of silver nitrate precipitate by using the 0.00350 moles and its molar mass (187.77g/mol):
[tex]m=0.00350molAgBr*\frac{187.77gAgBr}{1molAgBr}=0.657g[/tex]
And finally the percent yield:
[tex]Y=\frac{0.631g}{0.657g}*100\%\\\\Y=96.0\%[/tex]
Regards!
