Respuesta :
Answer: A. 5.41
Explanation:
To calculate the moles :
[tex]\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}[/tex]
[tex]\text{Moles of} Al=\frac{2.50g}{27g/mol}=0.0926moles[/tex]
[tex]\text{Moles of} Pb(NO_3)_2=\frac{8.65g}{331g/mol}=0.0261moles[/tex]
[tex]2Al(s)+3Pb(NO_3)_2(aq)\rightarrow 3Pb(s)+2Al(NO_3)_3(aq)[/tex]
According to stoichiometry :
3 moles of [tex]Pb(NO_3)_2[/tex] require = 2 moles of [tex]Al[/tex]
Thus 0.0261 moles of [tex]Pb(NO_3)_2[/tex] will require=[tex]\frac{2}{3}\times 0.0261=0.0174moles[/tex] of [tex]Al[/tex]
Thus [tex]Pb(NO_3)_2[/tex] is the limiting reagent as it limits the formation of product and [tex]Al[/tex] is the excess reagent.
As 3 moles of [tex]Pb(NO_3)_2[/tex] give = 3 moles of [tex]Pb[/tex]
Thus 0.0261 moles of [tex]Pb(NO_3)_2[/tex] give =[tex]\frac{3}{3}\times 0.0261=0.0261 moles[/tex] of [tex]Pb[/tex]
Mass of [tex]Pb=moles\times {\text {Molar mass}}=0.0261moles\times 207g/mol=5.41g[/tex]
Thus 5.41 g of solid lead will be produced from the given masses of both reactants.
