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weredQuestion 4Answer0/1 ptsCalculate the maximum mass (in g) of Pb(s) that canbe obtained from the reaction of 393. g PbS with593. g PbO. Enter your answer as an integer.2 PbO(s) + PbS(s) → 3 Pb(s) + SO₂(g)826 margin of error

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Explanation:

We are given: mass of PbS = 393g

: mass of PbO = 593g

We know: molar of PbS = 239.3g/mol

: molar mass of PbO = 223.2g/mol

: molar mass of Pb = 207.2g/mol

m is mass and M is molar mass.

We first determine the mass of Pb from the mass of PbS:

[tex]\begin{gathered} m(Pb)\text{ = }\frac{m(PbS)}{M(PbS)}\text{ }\times\frac{n(Pb)}{n(PbS)}\text{ }\times M(Pb) \\ \\ \text{ = }\frac{393}{239.3}\times\frac{3}{1}\times207.2 \\ \\ \text{ = 1020.85g} \end{gathered}[/tex]

We then determine the mass of Pb from the mass of PbO:

[tex]\begin{gathered} M(Pb)\text{ = }\frac{m(PbO)}{M(PbO)}\times\frac{n(Pb)}{n(PbO)}\times M(Pb) \\ \\ \text{ = }\frac{593}{223.2}\times\frac{3}{2}\times207.2 \\ \\ \text{ = 825.74g} \end{gathered}[/tex]

From the masses obtained above, it is clear to observe that PbO is the limiting reactant.

Answer:

mass of Pb = 826g

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