Answer:
Explanation:
The reaction that takes place is:
With a percent yield of 87.5%.
To determine the limiting reactant, first we convert the masses of each reactant to moles, using their molar mass:
Looking at the stoichiometric coefficients, we see that 1 mol of Pb(NO₃)₂ would react completely with 2 moles of KCl. Following that logic, 0.0296 mol Pb(NO₃)₂ would react completely with (2x0.0296) 0.0592 mol of KCl. We have more than that amount of KCl, this means KCl is the reactant in excess and Pb(NO₃)₂ is the limiting reactant.
To calculate the mass of precipitate (PbCl₂) formed, we use the moles of the limiting reactant:
- Keeping in mind the reaction yield, the moles of Pb(NO₃)₂ that would react are:
Now we convert that amount to moles of KCl and finally into grams of KCl:
3.86 g of KCl would react, so the amount remaining would be:
For the reaction between 9.82 g of lead(II) nitrate with 5.76 g of potassium chloride, we have:
a. The balanced chemical equation is:
Pb(NO₃)₂(aq) + 2KCl(aq) → PbCl₂(s) + 2KNO₃(aq)
b. The limiting reactant is lead(II) nitrate (Pb(NO₃)₂).
c. The mass of the precipitate (PbCl₂) formed is 7.20 g.
d. The mass of the excess reactant (KCl) that remains in solution after the reaction is 1.35 grams.
a. The chemical reaction between lead(II) nitrate and potassium chloride is the following:
Pb(NO₃)₂(aq) + KCl(aq) → PbCl₂(s) + KNO₃(aq) (1)
In this reaction, lead (II) chloride precipitates in the solution.
We need to balance equation (1). We can see that on the reactants side, we have 2 molecules of NO₃, and on the products side, we have one, so we need to add a coefficient of 2 before KNO₃
Pb(NO₃)₂(aq) + KCl(aq) → PbCl₂(s) + 2KNO₃(aq)
Now, we have 2 K atoms on the products, so we need to add a coefficient of 2 before KCl.
Pb(NO₃)₂(aq) + 2KCl(aq) → PbCl₂(s) + 2KNO₃(aq) (2)
Reaction (2) is balanced now.
b. To find the limiting reactant we need to calculate the initial number of moles of Pb(NO₃)₂ and KCl
[tex] n_{Pb(NO_{3})_{2}}_{i} = \frac{m_{Pb(NO_{3})_{2}}}{M_{Pb(NO_{3})_{2}}} [/tex]
Where:
m: is the mass
M: is the molar mass
[tex] n_{Pb(NO_{3})_{2}}_{i} = \frac{9.82 g}{331.2 g/mol} = 0.0296 \:moles [/tex]
[tex] n_{KCl}_{i} = \frac{m_{KCl}}{M_{KCl}} = \frac{5.76 g}{74.5513 g/mol} = 0.0773 \:moles [/tex]
From reaction (2), we have that 1 mol of Pb(NO₃)₂ react with 2 moles of KCl so, the number of moles of Pb(NO₃)₂ needed to react with 0.0773 moles of KCl is:
[tex] n_{Pb(NO_{3})_{2}} = \frac{1 \:mol \:Pb(NO_{3})_{2}}{2\: mol \:KCl}*n_{KCl}_{i} = \frac{1 \:mol \:Pb(NO_{3})_{2}}{2\: mol \:KCl}*0.0773 \:moles = 0.0387 \:moles [/tex]
Since we need 0.0387 moles of Pb(NO₃)₂ to react with KCl, and initially we have 0.0296 moles, the limiting reactant is Pb(NO₃)₂.
c. The mass of PbCl₂ can be calculated as follows.
From reaction (2) we have that 1 mol of Pb(NO₃)₂ produces 1 mol of PbCl₂, so:
[tex] n_{PbCl_{2}} = n_{Pb(NO_{3})_{2}}_{i} = 0.0296 \:moles [/tex]
[tex] m_{PbCl_{2}}_{t} = n_{PbCl_{2}}*M_{PbCl_{2}} = 0.0296 \:moles*278.1 g/mol = 8.23 g [/tex]
Since the percent yield is 87.5%, the mass of PbCl₂ formed (experimental mass) is:
[tex] m_{PbCl_{2}}_{e} = 8.23 g*\frac{87.5}{100} = 7.20 g [/tex]
Therefore, the mass of PbCl₂ formed is 7.20 g.
d. Mass of excess reactant
The mass of the excess reactant that remains in solution after the reaction is given by:
[tex] m_{KCl} = m_{KCl}_{i} - m_{KCl}_{r} [/tex]
The mass of KCl that reacts with Pb(NO₃)₂ ([tex]m_{KCl}_{r} [/tex]) can be calculated from the number of moles:
[tex] n_{KCl} = \frac{2\: mol \:KCl}{1 \:mol \:Pb(NO_{3})_{2}}*n_{Pb(NO_{3})_{2}}_{i} = \frac{2\: mol \:KCl}{1 \:mol \:Pb(NO_{3})_{2}}*0.0296 \:moles = 0.0592 \:moles [/tex]
[tex] m_{KCl}_{r} = n_{KCl}*M_{KCl} = 0.0592\:moles*74.5513 g/mol = 4.41 g [/tex]
Then, the mass of KCl that remains is:
[tex] m_{KCl} = m_{KCl}_{i} - m_{KCl}_{r} = 5.76 g - 4.41 g = 1.35 g [/tex]
Hence, 1.35 grams of KCl remains in solution after the reaction.
Find more here:
I hope it helps you!
