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:

              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

Answer 2

Answer: A)  5.41 g

Explanation:

To calculate the moles, we use the equation:

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

a) moles of $Pb(NO_3)_2$

$\text{Number of moles}=\frac{8.65g}{331.2g/mol}=0.0261moles$

b) moles of $Al$

$\text{Number of moles}=\frac{2.50g}{26.9g/mol}=0.0930moles$

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=$\frac{2}{3}\times 0.0261=0.0174moles$  of $Al$

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

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

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

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

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