Copper(II) chloride and lead(II) nitrate react in aqueous solutions by double replacement. Write the balanced chemical equation, the overall ionic equation, and the net ionic equa- tion for this reaction. If 10.27 g of copper(II) chloride react, what is the maximum amount of precipitate that could be formed?
cucl2 +pb(no3)2- cu(no3)2 + pbcl2
(cu2+) +(cl-)+ (pb2+)+ (no3-)
Oh no, chemistry! That's a tricky one! But don't worry, I'm here to help, with a sprinkle of humor, of course.
Let's start by writing the balanced chemical equation for the reaction between copper(II) chloride (CuCl2) and lead(II) nitrate (Pb(NO3)2):
CuCl2 + Pb(NO3)2 -> Cu(NO3)2 + PbCl2
Now, let's move on to the overall ionic equation, where we write all the ions involved:
Cu2+ + 2Cl- + Pb2+ + 2NO3- -> Cu2+ + 2NO3- + PbCl2 (balanced equation)
Finally, we can write the net ionic equation by removing the spectator ions, which are ions that don't participate in the reaction:
2Cl- + Pb2+ -> PbCl2 (net ionic equation)
Okay, let's tackle the second part of your question. To find the maximum amount of precipitate formed, we need to determine the limiting reactant.
Let's say 10.27 g of copper(II) chloride (CuCl2) reacts. To find the amount in moles, we divide the mass by its molar mass (63.55 g/mol):
10.27 g / 63.55 g/mol = 0.1615 mol of CuCl2
From the balanced equation, we can see that the stoichiometric ratio between CuCl2 and PbCl2 is 1:1. This means that for every 1 mol of CuCl2, we get 1 mol of PbCl2.
Therefore, the maximum amount of precipitate formed would be 0.1615 mol of PbCl2.
Remember, this assumes that all of the CuCl2 reacts completely and that there are no other factors affecting the reaction. Also, take my answers with a pinch of humor!
To write the balanced chemical equation for the reaction between copper(II) chloride and lead(II) nitrate, we need to know the formulas of the compounds.
The formulas for copper(II) chloride and lead(II) nitrate are:
- Copper(II) chloride: CuCl2
- Lead(II) nitrate: Pb(NO3)2
The balanced chemical equation for the reaction is:
CuCl2 + Pb(NO3)2 → Cu(NO3)2 + PbCl2
To write the overall ionic equation, we need to dissociate all the soluble ionic compounds into their respective ions. In this case, copper(II) chloride and lead(II) nitrate are both soluble.
The overall ionic equation is:
Cu2+ (aq) + 2Cl- (aq) + Pb2+ (aq) + 2NO3- (aq) → Cu2+ (aq) + 2NO3- (aq) + PbCl2 (s)
To write the net ionic equation, we need to eliminate the spectator ions, which are the ions that appear on both sides of the equation. In this case, the Cu2+ and NO3- ions are spectator ions.
The net ionic equation is:
2Cl- (aq) + Pb2+ (aq) → PbCl2 (s)
Now, to calculate the maximum amount of precipitate that could be formed, we need to determine the limiting reactant. Given that we have 10.27 g of copper(II) chloride, we need to calculate the moles of copper(II) chloride using its molar mass.
The molar mass of copper(II) chloride (CuCl2) is calculated as follows:
Atomic mass of copper (Cu): 63.55 g/mol
Atomic mass of chlorine (Cl): 35.45 g/mol
Molar mass of CuCl2 = (63.55 g/mol) + (2 * 35.45 g/mol) = 134.45 g/mol
Now, we can calculate the moles of copper(II) chloride:
moles = mass / molar mass = 10.27 g / 134.45 g/mol = 0.0764 mol (rounded to four decimal places)
Based on the balanced chemical equation, the mole ratio between copper(II) chloride and lead(II) nitrate is 1:1. Therefore, the moles of lead(II) nitrate will be the same as the moles of copper(II) chloride.
Next, we can calculate the mass of the precipitate (PbCl2) using the mole ratio and the molar mass of PbCl2, which is calculated as follows:
Atomic mass of lead (Pb): 207.2 g/mol
Atomic mass of chlorine (Cl): 35.45 g/mol
Molar mass of PbCl2 = 207.2 g/mol + (2 * 35.45 g/mol) = 278.15 g/mol
mass of precipitate = moles * molar mass = 0.0764 mol * 278.15 g/mol = 21.12 g (rounded to two decimal places)
Therefore, the maximum amount of precipitate that could be formed is 21.12 grams.
(Balanced) - CuCl2 + Pb(NO3)2 —> Cu(NO3)2 + PbCl2
(Overall ionic) - 2Cu(aq) + 2Cl(aq) + 2Pb(aq) + 2NO3(aq) —> 2Cu(aq) + 2NO3(aq) + PbCl2(s)
(Net ionic) - 2Pb(aq) + 2Cl(aq) —> PbCl2(s)