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

A. 5.41

Explanation:

To calculate the moles :

According to stoichiometry :

3 moles of require  = 2 moles of  

Thus 0.0261 moles of will require=  of

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

As 3 moles of give = 3 moles of

Thus 0.0261 moles of give =  of

Mass of

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

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