When 496.5 grams of Pb(NO3)2 reacts completely with KBr, how much will the
total mass of the products be? Explain your answer.
Mass mass problem - mass of reactant to mass of product
The total mass of the products will be 576.5 grams. This is because the reaction of 496.5 grams of Pb(NO3)2 with KBr produces 80 grams of PbBr2 and 496.5 grams of KNO3, for a total mass of 576.5 grams.
To determine the total mass of the products, we first need to find the molar masses of Pb(NO3)2 and KBr.
The molar mass of Pb(NO3)2 can be calculated by adding the molar masses of the individual elements:
Pb (lead) = 207.2 g/mol
N (nitrogen) = 14.01 g/mol
O (oxygen) = 16.00 g/mol (there are 3 oxygen atoms in each nitrate molecule)
So, the molar mass of Pb(NO3)2 = (207.2 + 14.01 + (16.00 x 3)) g/mol = 331.2 g/mol.
The molar mass of KBr can be determined by adding the molar masses of potassium (K) and bromine (Br):
K (potassium) = 39.10 g/mol
Br (bromine) = 79.90 g/mol
So, the molar mass of KBr = (39.10 + 79.90) g/mol = 119.0 g/mol.
Next, we need to find the moles of Pb(NO3)2 in 496.5 grams using the molar mass:
Moles of Pb(NO3)2 = mass (g) / molar mass (g/mol) = 496.5 g / 331.2 g/mol ≈ 1.50 mol.
According to the balanced chemical equation, 1 mole of Pb(NO3)2 reacts with 2 moles of KBr to produce 2 moles of KNO3 and 1 mole of PbBr2.
Since the stoichiometry of the reaction implies that the number of moles of the reactant and the product is the same, we can conclude that the number of moles of PbBr2 formed will also be approximately 1.50 mol.
Finally, we can find the total mass of the products:
Total mass of products = moles of PbBr2 x molar mass of PbBr2
= 1.50 mol x (207.2 g/mol + (79.90 x 2) g/mol)
= 1.50 mol x 367.0 g/mol
= 550.5 grams.
Therefore, the total mass of the products formed when 496.5 grams of Pb(NO3)2 reacts completely with KBr will be approximately 550.5 grams.