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?

Question

Wyatt Atkins · Accepted Answer

See explanation.

The use of the limiting reactant and stoichiometry are the topics of this query. The response 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 this is of use to you.

Cameron Andrews · Answer

You should have more of one reactant than you require in relation to the other, as I would anticipate in advance.

In actuality, the amounts of each product you can obtain are limited to what you can use of one of the reactants before it runs out completely.

The limiting reactant is the one that is used up initially.

You can't really tell which is which, so let's just guess. Zinc is my first guess.

We require:

#"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.

In reality, we possess:

#"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)#

Yes, we need to use zinc, the limiting reagent, to calculate the amounts of each product because we have more lead nitrate than we need. This is essentially an extension of the stoichiometric conversion shown in the first calculation.

#"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)#

Of course, we follow a similar procedure for the other product. Alright, let's practice a little bit by repeating a few steps.

#"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.

Avery Adams · Answer

undefined To determine the masses of each product generated from the reaction, we first need to write and balance the chemical equation for the reaction between zinc (Zn) and lead(II) nitrate (Pb(NO3)2):

Zn + Pb(NO3)2 → Zn(NO3)2 + Pb

Next, we need to calculate the limiting reactant by finding out which reactant is completely consumed first. This can be done by calculating the number of moles of each reactant using their respective molar masses and then comparing the mole ratios from the balanced equation.

For zinc (Zn):
- Molar mass of Zn = 65.38 g/mol
- Number of moles of Zn = mass / molar mass = 1.3 g / 65.38 g/mol ≈ 0.0199 mol

For lead(II) nitrate (Pb(NO3)2):
- Molar mass of Pb(NO3)2 = 331.21 g/mol
- Number of moles of Pb(NO3)2 = mass / molar mass = 8.28 g / 331.21 g/mol ≈ 0.0250 mol

According to the balanced equation, the stoichiometric ratio between Zn and Pb(NO3)2 is 1:1. Since zinc (Zn) has the smaller number of moles, it is the limiting reactant.

Now, we can use the stoichiometry of the balanced equation to determine the masses of the products:

For zinc nitrate (Zn(NO3)2):
- According to the balanced equation, 1 mole of Zn(NO3)2 is produced for every mole of Zn consumed.
- Mass of Zn(NO3)2 = number of moles of Zn × molar mass of Zn(NO3)2
- Mass of Zn(NO3)2 = 0.0199 mol × (65.38 + 3(16.00) + 2(14.01)) g/mol ≈ 2.54 g

For lead (Pb):
- According to the balanced equation, 1 mole of Pb is produced for every mole of Zn consumed.
- Mass of Pb = number of moles of Zn × molar mass of Pb
- Mass of Pb = 0.0199 mol × 207.2 g/mol ≈ 4.12 g

Therefore, approximately 2.54 grams of zinc nitrate (Zn(NO3)2) and 4.12 grams of lead (Pb) would be generated from the reaction.