STOICHIOMETRY REACTION

Question

How many grams of PbBr₂ will precipitate when excess FeBr₂ solution is added to 43.0 mL of 0.788 M Pb(NO₃)₂ solution?

Pb(NO₃)₂(aq) + FeBr₂(aq) PbBr₂(s) + Fe(NO₃)₂(aq)

Answer 1

Hello, Raphael,

The first step is to check that the equation is balanced. It is, so I know that for every 1 mole of Pb(NO3)2 in the reactants, we'll get one mole of the PbBr2. We have excess FeBr2, so that means that we only need to know how many moles of Pb(NO3)2 we are adding, since the same number of moles of PBr2 will be produced. We'll assume it all precipitates and that the reaction goes to completion.

Calculate the number of moles of Pb(NO3)2 by first calculating the molar mass (by adding up the individual atoms), which I find to be 331.2 grams/mole. We are given 43.0 ml of a 0.788 M Pb(NO3)2 solution, so let's calculate the total moles of Pb(NO3)2 added.

(We are limited to 3 significant digits since both 43.0 and 0.788 are both 3 sig figs)

A M(olar) solution is defined as the number of moles per liter, or moles/L. So we can restate 0.788 M as 0.788 moles/Liter. Since we are given a volume in ml, we need to covert it to liters, first.

Since 1000 ml = 1 Liter, we can calculate the number of moles as follows:

(43 ml)*(1 L/1000 ml)*(0.788 moles/L)

Now this is the fun part: Check to see if the units cancel to give us what we want: just moles.

The ml cancels out, L cancels out, and all we're left with is moles Pb(NO3)2. A good sign.

The numbers calculate to 0.0339 moles of Pb(NO3)2. The balanced equation tells us we should get the same amount of PbBr2 on the other side.

Aha! Since the FeBr2 is in excess, this means that if we start with 0.0339 moles of Pb(NO3)2, we'll get 0.0339 moles of PbBr2. Close, . . .

but the question wants to know the number of "grams" of PbBr2. We can calculate that with knowing the molar mass of PBr2, which is 367 g/mole.

Therefore,

0.0339 moles PBr2 * 367 g/mole

should give us what we need. Cancel the mole units and we're left with just grams, so that means we are using the conversion factor (the molar mass) in the correct manner. The answer is therefore: 12.4 g of PbBr2.

[Note: the molar mass is basically a conversion unit. It tells us that 367 g of PbBr2 makes one mole of the compound. We can also read it as one mole of PbBr2 is equal to 367 grams. So we can invert the value if we want, which would put the unit of moles on top and grams on bottom.

So molar mass can be expressed as both 367 g/mole and as (1 mole/367 grams). If you had an experimental value expressed in grams and wanted to find the number of moles, multiply it by this form of the molar mass expression. The grams cancel and you are left with moles.

I hope this helps,

Bob

Answer 2

Pb(NO₃)₂(aq) + FeBr₂(aq) PbBr₂(s) + Fe(NO₃)₂(aq) ... balanced equation

moles Pb(NO₃)₂ present = 43.0 ml x 1 L/1000 ml x 0.788 mol/L = 0.033884 moles

moles PbBr₂ formed = 0.033884 mol Pb(NO₃)₂ x 1 mol PbBr₂/mol Pb(NO₃)₂ = 0.0033884 moles

Grams PbBr₂ formed = 0.0033884 mol x 367 g/mol = 12.4 g (3 sig. figs.)