Question

use the chemical equation and the table to answer the question.
pb(no₃)₂(aq) + 2kbr(aq) → pbbr₂(s) + 2kno₃(aq)
molar mass (g/mol) of reactants and
products involved in a reaction
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)
550.5 g
853.5 g
496.5 g
702.0 g

Explanation:

Step1: Calculate the moles of \(Pb(NO_3)_2\)

The molar mass of \(Pb(NO_3)_2\) is \(331\space g/mol\).
Using the formula \(n=\frac{m}{M}\) (where \(n\) is moles, \(m\) is mass, \(M\) is molar mass), for \(Pb(NO_3)_2\) with \(m = 496.5\space g\) and \(M=331\space g/mol\), we have \(n=\frac{496.5}{331}=1.5\space mol\).

Step2: Determine the moles of \(KBr\) using stoichiometry

From the balanced equation \(Pb(NO_3)_2(aq)+2KBr(aq)\to PbBr_2(s)+2KNO_3(aq)\), the mole ratio of \(Pb(NO_3)_2\) to \(KBr\) is \(1:2\).
So, moles of \(KBr = 2\times1.5 = 3\space mol\).

Step3: Calculate the mass of \(KBr\)

The molar mass of \(KBr\) is \(119\space g/mol\).
Using \(m = n\times M\), mass of \(KBr=3\times119 = 357\space g\).

Step4: Apply the law of conservation of mass

According to the law of conservation of mass, mass of reactants = mass of products.
Mass of reactants \(=496.5 + 357=853.5\space g\).

Answer:

\(853.5\space g\)