When 0.25M (Molarity) aqueous solutions of ammonium chloride and lead (II) nitrate are mixed a precipitate forms. How many grams of solid would be produced by mixing 250 mL of each solution?
This is a limiting reagent problem since BOTH reactants are given.
2NH4Cl + Pb(NO3)2 ==> PbCl2(s) + 2NH4NO3
1.Calculate moles NH4Cl = M x L = ??
2.Calculate moles Pb(NO3)2 = M x L = ??
3a. Using the coefficients in the balanced equation, convert moles NH4Cl to moles of the product.
3b. Same procedure, convert moles Pb(NO3)2 to moles of the product.
3c. It is quite likely, since this is a limiting reagent problem, that the answers from 3a and 3b will not agree which means one of them is wrong. The correct answer, in limiting reagent problems, is ALWAYS the smaller value and the reagent producing that value is the limiting reagent.
4. Convert moles from 3c to grams. g = moles x molar mass.
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To determine the number of grams of solid produced by mixing the solutions, we first need to find the limiting reactant. The limiting reactant is the reactant that will be completely consumed and determines the maximum amount of product that can be formed.
To find the limiting reactant, we need to compare the number of moles of each reactant. The equation for the reaction between ammonium chloride (NH4Cl) and lead (II) nitrate (Pb(NO3)2) is:
2NH4Cl + Pb(NO3)2 → 2NH4NO3 + PbCl2
First, let's calculate the number of moles of ammonium chloride:
Molarity (M) = moles/volume (L)
Given:
Molarity of ammonium chloride = 0.25 M
Volume of ammonium chloride solution = 250 mL = 0.25 L
Number of moles of ammonium chloride = Molarity × volume
= 0.25 M × 0.25 L
= 0.0625 moles
Next, we calculate the number of moles of lead (II) nitrate. Since the stoichiometric coefficient of Pb(NO3)2 is 1, the number of moles of lead (II) nitrate will be equal to the molarity:
Number of moles of lead (II) nitrate = Molarity = 0.25 moles
To determine the limiting reactant, we compare the moles of each reactant. In this case, both reactants have the same number of moles. Therefore, neither reactant is in excess, and both will be completely consumed.
The balanced equation tells us that for every 2 moles of ammonium chloride, 1 mole of lead chloride (PbCl2) is produced. Hence, the ratio of ammonium chloride to lead chloride is 2:1.
Since there are 0.0625 moles of ammonium chloride, there will be half as many moles of lead chloride formed (0.0625/2 = 0.03125 moles).
Finally, to calculate the mass of lead chloride formed, we need to know the molar mass of PbCl2. The molar mass of lead chloride can be calculated by adding the atomic masses of lead (Pb) and chlorine (Cl) in the compound.
Atomic mass of Pb = 207.2 g/mol
Atomic mass of Cl = 35.5 g/mol
Molar mass of PbCl2 = (207.2 g/mol) + 2 × (35.5 g/mol)
= 207.2 g/mol + 71 g/mol
= 278.2 g/mol
Now, we can calculate the mass of lead chloride formed:
Mass of lead chloride = Moles × Molar mass
= 0.03125 moles × 278.2 g/mol
= 8.69 grams
Therefore, by mixing 250 mL of each solution, approximately 8.69 grams of lead chloride (PbCl2) will be produced.