Question

You dissolve 8.65 grams of lead(II) nitrate in water, and then you add 2.50 grams of aluminum. This reaction occurs: 2Al(s) + 3Pb(NO3)2(aq) → 3Pb(s) + 2Al(NO3)3(aq). What’s the theoretical yield of solid lead? Use the ideal gas resource and the periodic table. A. 5.41 g B. 11.2 g C. 19.2 g D. 28.8 g Reset

Answer 1

Answer: A. 5.41

Explanation:

To calculate the moles :

$\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}$    

$\text{Moles of} Al=\frac{2.50g}{27g/mol}=0.0926moles$

$\text{Moles of} Pb(NO_3)_2=\frac{8.65g}{331g/mol}=0.0261moles$

$2Al(s)+3Pb(NO_3)_2(aq)\rightarrow 3Pb(s)+2Al(NO_3)_3(aq)$  

According to stoichiometry :

3 moles of $Pb(NO_3)_2$ require  = 2 moles of $Al$  

Thus 0.0261 moles of $Pb(NO_3)_2$ will require=$\frac{2}{3}\times 0.0261=0.0174moles$  of $Al$

Thus $Pb(NO_3)_2$ is the limiting reagent as it limits the formation of product and $Al$ is the excess reagent.

As 3 moles of $Pb(NO_3)_2$ give = 3 moles of $Pb$

Thus 0.0261 moles of $Pb(NO_3)_2$ give =$\frac{3}{3}\times 0.0261=0.0261 moles$  of $Pb$

Mass of $Pb=moles\times {\text {Molar mass}}=0.0261moles\times 207g/mol=5.41g$

Thus 5.41 g of solid lead will be produced from the given masses of both reactants.