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

Answer 1

Answer: A) 5.41 g

Step-by-step explanation:

To calculate the moles, we use the equation:

`\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}`

a) moles of
`Pb(NO_3)_2`

`\text{Number of moles}=(8.65g)/(331.2g/mol)=0.0261moles`

b) moles of
`Al`

`\text{Number of moles}=(2.50g)/(26.9g/mol)=0.0930moles`

The balanced reaction is:

`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` 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.

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=0.0261moles` of
`Pb`

Mass of
`Pb=moles* {\text {Molar mass}}=0.0261moles* 207.2g/mol=5.41g`

Thus 5.41 g of
`Pb` will be produced from the given masses of both reactants.

Answer 2

Option-A = 5.41 g of Pb

Solution:

The Balance Chemical Equation is as follow,

2 Al + 3 Pb(NO₃)₂ → 3 Pb + 2 Al(NO₃)₃

Step 1: Calculate the Limiting Reagent,

According to Balance equation,

53.96 g (2 mol) Al reacts with = 993.6 g (3 mol) of Pb(NO₃)₂

So,

2.50 g of Al will react with = X g of Pb(NO₃)₂

Solving for X,

X = (2.50 g × 993.6 g) ÷ 53.96 g

X = 46.03 g of Pb(NO₃)₂

It means 2.50 g of Aluminium requires 46.03 g of Pb(NO₃)₂, while we are provided with only 8.65 g of Pb(NO₃)₂. Therefore, Pb(NO₃)₂ is the limiting reagent and will control the yield of products.

Step 2: Calculate amount of Lead produced,

According to equation,

993.6 g (3 mol) of Pb(NO₃)₂ produces = 621.6 g of Pb

So,

8.65 g (3 mol) of Pb(NO₃)₂ will produce = X g of Pb

Solving for X,

X = (8.65 g × 621.6 g) ÷ 993.6 g

X = 5.41 g of Pb