Use the chemical equation and the table to answer the question. Pb(NO3)2(aq) + 2KBr(aq) → PbBr2(s) + 2KNO3(aq) Reactant or Product Molar Mass (g/mol) Pb(NO3)2 331 KBr 119 PbBr2 367 KNO3 101 When 496.5 grams of Pb(NO3)2 reacts completely with KBr, how much will the total mass of the products be? (1 point) Responses 702.0 g 702.0 g 496.5 g 496.5 g 853.5 g 853.5 g 550.5 g 550.5 g
To find the total mass of the products, we first need to calculate the moles of Pb(NO3)2 using its molar mass:
496.5 g Pb(NO3)2 x (1 mol Pb(NO3)2/331 g Pb(NO3)2) = 1.5 mol Pb(NO3)2
From the balanced chemical equation, we see that 1 mol of Pb(NO3)2 produces 1 mol of PbBr2. So, 1.5 mol Pb(NO3)2 will produce 1.5 mol PbBr2.
Now, we can calculate the mass of PbBr2:
1.5 mol PbBr2 x 367 g PbBr2/mol PbBr2 = 550.5 g PbBr2
The total mass of the products will be the mass of PbBr2 plus the mass of KNO3 produced (since it's in a 2:1 ratio with Pb(NO3)2):
550.5 g PbBr2 + 2(1.5 mol KNO3 x 101 g KNO3/mol KNO3) = 853.5 g
Therefore, the correct answer is 853.5 g.