Answer:
Explanation:
In this case, we need to write the reaction that is taking place here. We have lead (II) chloride and silver nitrate:
PbCl₂ + 2AgNO₃ ----------> Pb(NO₃)₂ + 2AgCl
According to this balanced reaction, 1 mole of PbCl₂ reacts with 2 moles of AgNO₃, which mean that we have a mole ratio 1:2. Based on this, we can determine which is the limiting reactant. Let's calculate first the moles of each reactant with the given masses using the following expression:
n = mass/MM (1) MM: molecular weight
To get the molecular weight, we need either the atomic weights of the elements that form the compound, or looking out in handbook. Now, let's write the atomic weight of each element:
Ag: 107.87 g/mol; N: 14 g/mol; O: 16 g/mol; Cl: 35.5 g/mol; Pb: 207.2 g/mol
Molecular weights:
MMPbCl₂ = 207.2 + (35.5 * 2) = 278.2 g/mol
MMAgNO₃ = 107.87 + 14 + (3*16) = 169.87 g/mol
MMAgCl = 107.87 + 35.5 = 143.37 g/mol
The moles of the reactants are:
nPbCl₂ = 42.7 / 278.2 = 0.1535 moles
nAgNO₃ = 93.2 / 169.87 = 0.5487 moles
We have the moles, now let's compare it to the mole ratio of above to determine the limiting reagent:
If: 1 mole PbCL₂ reacts with 2 moles of AgNO₃
Then: 0.1535 moles PbCl₂ reacts with X moles of AgNO₃
X = 0.1535 * 2 / 1
X = 0.3070 moles of AgNO₃
As we can see, we only need 0.3070 moles of AgNO₃ to completely reacts with the moles of lead chloride. However, we have more of that, we have 0.5487 moles. This means that the AgNO₃ has an excess of reactant, hence, the limiting reactant is the PbCl₂.
Now that we know who's the limiting, we can determine the amount of AgCl.
According to the balanced reaction, 1 mole of PbCl₂ reacts and produce 2 moles of AgCl, so, the moles produced will be doubled in the reaction, hence:
nAgCl = 0.1535 * 2 = 0.3070 moles
Finally, we can determine the amount of this compound, using the molecular mass of AgCl, which we already have:
m = 0.3070 * 143.37
<h2>
mAgCl = 44.0146 g</h2>
Hope this helps
