What masses of each product would be generated from the reaction of #"1.3 g"# of zinc with #"8.28 g"# of lead(II) nitrate?

2 Answers
Dec 4, 2015

Answer:

See explanation.

Explanation:

This question is about stoichiometry and the use of the limiting reactant.

The reaction is:

#AB+C->AC+B#

The molar ratio between #AB# and #C# is #1:1#. Therefore, theoretically #1mol# of #AB# will react with #1mol# of #C#.

Experimentally you have #0.025mol# of #AB# and #0.02mol# of #C#.

To consume all of the #C# you will need only #0.02mol# of #AB#, which means that you have excess of #AB=0.025-0.02=0.005mol#

Therefore, #C# is the limiting reactant and its number of mole will be used to calculate the amounts of the products formed.

I hope you find this helpful.

Dec 4, 2015

I would expect ahead of time that you have more of one reactant than you need in comparison to the other.

You really can only get amounts of each product proportional to the amount of one of the reactants that you manage to use before it's all gone.

This reactant that is used up first is the Limiting Reactant.

Since you don't really know which one is which, let's just figure it out. My first guess is zinc.

We need:

#"mol Pb"("NO"_3)_2#

#= "1.30" cancel"g Zn" xx cancel"1 mol Zn"/("65.38" cancel"g Zn") xx ("1 mol Pb"("NO"_3)_2)/cancel"1 mol Zn"#

#= color(green)("0.01988 mol Pb"("NO"_3)_2")#

But, let's check how many #"mol"#s of this we actually have available.

We actually have:

#"mol Pb"("NO"_3)_2#

#= "8.28" cancel("g Pb"("NO"_3)_2) xx ("1 mol Pb"("NO"_3)_2)/("331.2074" cancel("g Pb"("NO"_3)_2))#

#= color(green)("0.025 mol Pb"("NO"_3)_2)#

Yeah, we have more than we need of lead nitrate, so we have to use zinc, the limiting reagent, to determine the amounts of each product. (It's more of the same stoichiometric conversion you see in the first calculation, but extended one more step.)

#"g Zn"("NO"_3)_2#

#= "1.30" cancel"g Zn" xx cancel"1 mol Zn"/("65.38" cancel"g Zn") xx cancel("1 mol Zn"("NO"_3)_2)/cancel"1 mol Zn" #

# xx ("189.387" cancel("g Zn"("NO"_3)_2))/cancel("1 mol Zn"("NO"_3)_2)#

#"= "color(blue)("3.77 g Zn"("NO"_3)_2)#

Naturally, we do something similar for the other product. And let's repeat some steps for practice, whaddya say?

#"g Pb"#

#= "1.30" cancel"g Zn" xx cancel"1 mol Zn"/("65.38" cancel"g Zn") xx cancel("1 mol Pb")/cancel"1 mol Zn" #

# xx ("207.2 g Pb")/cancel("1 mol Pb")#

#"= "color(blue)("4.12 g Pb")#

I'm assuming that you meant to write #"1.30 g"#, not #"1.3 g"#, as your other mass has 3 sig figs.