Question

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

Answer 1

Step-by-step 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:

`\begin{gathered} m(Pb)\text{ = }(m(PbS))/(M(PbS))\text{ }*(n(Pb))/(n(PbS))\text{ }* M(Pb) \\ \\ \text{ = }(393)/(239.3)*(3)/(1)*207.2 \\ \\ \text{ = 1020.85g} \end{gathered}`

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

`\begin{gathered} M(Pb)\text{ = }(m(PbO))/(M(PbO))*(n(Pb))/(n(PbO))* M(Pb) \\ \\ \text{ = }(593)/(223.2)*(3)/(2)*207.2 \\ \\ \text{ = 825.74g} \end{gathered}`

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

Answer:

mass of Pb = 826g