Problem 1:
Solution (c):
We carry out two sep-a-rate calculations. First, starting with 36.8 g of Zn, we calculate the number of moles of ZnS that could be produced if all the Zn reacted according to the following conversions
grams of Zn → moles of Zn → moles of ZnS
Combining the conversions in one step, we write
[tex]moles \: of \: ZnS = 36.8 \: g \: Zn \times \frac{1 \: mol \: Zn}{65.378 \: g \: Zn} \times \frac{1 \: mol \: ZnS}{1 \: mol \: Zn}[/tex]
[tex]moles \: of \: ZnS = 0.56288 \: mol \: ZnS[/tex]
Second, for 19.4 g of S, the conversions are
grams of S → moles of S → moles of ZnS
The number of moles of ZnS that could be produced if all the S reacted is
[tex]moles \: of \: ZnS = 19.4 \: g \: S \times \frac{1 \: mol \: S}{32.065 \: g \: S} \times \frac{1 \: mol \: ZnS}{1 \: mol \: S}[/tex]
[tex]moles \: of \: ZnS = 0.60502 \: mol \: ZnS[/tex]
It follows that
Zn must be the limiting reageant because it produces a smaller amount of ZnS.
Solution (a):
The number of moles of ZnS that could be produced if all the S reacted is
[tex]moles \: of \: ZnS = 19.4 \: g \: S \times \frac{1 \: mol \: S}{32.065 \: g \: S} \times \frac{1 \: mol \: ZnS}{1 \: mol \: S}[/tex]
[tex]\blue{moles \: of \: ZnS = 0.605 \: mol \: ZnS}[/tex]
Solution (b):
The mass of ZnS that could be produced if all the Zn reacted is
[tex]mass \: of \: ZnS = 36.8 \: g \: Zn \times \frac{1 \: mol \: Zn}{65.378 \: g \: Zn} \times \frac{1 \: mol \: ZnS}{1 \: mol \: Zn} \times \frac{97.443 \: g \: ZnS}{1 \: mol \: ZnS}[/tex]
[tex]\blue{mass \: of \: ZnS = 54.8 \: g \: ZnS}[/tex]
Solution (d):
Starting with 0.56288 mol of ZnS, we can determine the mass of S that reacted using the mole ratio from the balanced equation and the molar mass of S. The conversion steps are:
moles of ZnS → moles of S → grams of S
so that
[tex]mass \: of \: S \: reacted = 0.56288 \: mol \: ZnS \times \frac{1 \: mol \: S}{1 \: mol \: ZnS} \times \frac{32.065 \: g \: S}{1 \: mol \: S}[/tex]
[tex]mass \: of \: S \: reacted = 18.049 \: g \: S[/tex]
The amount of S remaining (in excess) is the difference between the initial amount (19.4 g) and the amount reacted (18.049 g):
mass of S remaining = 19.4 g - 18.049 g = 1.35 g
Problem 2:
Solution (c):
We carry out two sep-a-rate calculations. First, starting with 126.4 g of NaOH, we calculate the number of moles of NaAlO₂ that could be produced if all the NaOH reacted according to the following conversions
grams of NaOH → moles of NaOH → moles of NaAlO₂
Combining the conversions in one step, we write
[tex]moles \: of \: NaAlO_{2} = 126.4 \: g \: NaOH \times \frac{1 \: mol \: NaOH}{39.9971 \: g \: NaOH} \times \frac{2 \: mol \: NaAlO_{2}}{2 \: mol \: NaOH}[/tex]
[tex]moles \: of \: NaAlO_{2} = 3.16023 \: mol \: NaAlO_{2}[/tex]
Second, for 97.70 g of Al, the conversions are
grams of Al → moles of Al → moles of NaAlO₂
The number of moles of NaAlO₂ that could be produced if all the Al reacted is
[tex]moles \: of \: NaAlO_{2} = 97.70 \: g \: Al \times \frac{1 \: mol \: Al}{26.9815 \: g \: Al} \times \frac{2 \: mol \: NaAlO_{2}}{2 \: mol \: Al}[/tex]
[tex]moles \: of \: NaAlO_{2} = 3.621 \: mol \: NaAlO_{2}[/tex]
It follows that
NaOH must be the limiting reageant because it produces a smaller amount of NaAlO₂.
Solution (a):
The number of moles of NaAlO₂ that could be produced if all the NaOH reacted is
[tex]moles \: of \: NaAlO_{2} = 126.4 \: g \: NaOH \times \frac{1 \: mol \: NaOH}{39.9971 \: g \: NaOH} \times \frac{2 \: mol \: NaAlO_{2}}{2 \: mol \: NaOH}[/tex]
[tex]\blue{moles \: of \: NaAlO_{2} = 3.160 \: mol \: NaAlO_{2}}[/tex]
Solution (b):
The mass of NaAlO₂ that could be produced if all the NaOH reacted is
[tex]mass \: of \: NaAlO_{2} = 126.4 \: g \: NaOH \times \frac{1 \: mol \: NaOH}{39.9971 \: g \: NaOH} \times \frac{2 \: mol \: NaAlO_{2}}{2 \: mol \: NaOH} \times \frac{81.9701 \: g \: NaAlO_{2}}{1 \: mol \: NaAlO_{2}}[/tex]
[tex]\blue{mass \: of \: NaAlO_{2} = 259.0 \: g \: NaAlO_{2}}[/tex]
Solution (d):
Starting with 3.16023 mol of NaAlO₂, we can determine the mass of Al that reacted using the mole ratio from the balanced equation and the molar mass of Al. The conversion steps are:
moles of NaAlO₂ → moles of Al → grams of Al
so that
[tex]mass \: of \: Al \: reacted = 3.16023 \: mol \: NaAlO_{2} \times \frac{2 \: mol \: Al}{2 \: mol \: NaAlO_{2}} \times \frac{26.9815 \: g \: Al}{1 \: mol \: Al}[/tex]
[tex]mass \: of \: Al \: reacted = 85.27 \: g \: Al[/tex]
The amount of Al remaining (in excess) is the difference between the initial amount (97.70 g) and the amount reacted (85.27 g):
mass of Al remaining = 97.70 g - 85.27 g = 12.43 g
#CarryOnLearning
