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

              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

A)  5.41 g

Explanation:

To calculate the moles, we use the equation:

a) moles of

b) moles of

The balanced reaction is:

According to stoichiometry :

3 moles of require 2 moles of

Thus 0.0261 moles of require=  of

Thus is the limiting reagent as it limits the formation of product.

As 3 moles of give = 3 moles of

Thus 0.0261 moles of give =  of

Mass of

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

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