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

Step-by-step explanation:

To calculate the moles :

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

`\text{Moles of} Al=(2.50g)/(27g/mol)=0.0926moles`

`\text{Moles of} Pb(NO_3)_2=(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=
`(2)/(3)* 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 =
`(3)/(3)* 0.0261=0.0261 moles` of
`Pb`

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

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