Copper(II) chloride and lead(II) nitrate react in aqueous solutions by double replacement. Write the balanced chemical equation, the over all ionic equation, and the net ionic equation for this reaction. If 16.06g of copper(II) chloride react, what is the maximum amount of precipitate that could be formed?
Answer in units of g.
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To write the balanced chemical equation, we need to determine the formulas and charges of the ions involved.
Copper(II) chloride is made up of copper ions with a charge of +2 (Cu²⁺) and chloride ions with a charge of -1 (Cl⁻).
Lead(II) nitrate is made up of lead ions with a charge of +2 (Pb²⁺) and nitrate ions with a charge of -1 (NO₃⁻).
The balanced chemical equation for the reaction can be written as follows:
CuCl₂ + Pb(NO₃)₂ → Cu(NO₃)₂ + PbCl₂
To write the overall ionic equation, we need to separate the soluble compounds into their respective ions. The compounds that are soluble in water are dissociated into ions.
CuCl₂(aq) + Pb(NO₃)₂(aq) → Cu(NO₃)₂(aq) + PbCl₂(s)
To write the net ionic equation, we eliminate the spectator ions, which are the ions that are present on both sides of the equation and do not participate in the reaction.
Cu²⁺(aq) + 2Cl⁻(aq) + Pb²⁺(aq) + 2NO₃⁻(aq) → Cu²⁺(aq) + 2NO₃⁻(aq) + PbCl₂(s)
By canceling out the spectator ions, we get the net ionic equation:
2Cl⁻(aq) + Pb²⁺(aq) → PbCl₂(s)
Now, to calculate the maximum amount of precipitate formed, we need to determine the stoichiometry of the reaction. From the balanced chemical equation, we can see that for every 1 mole of CuCl₂, 1 mole of PbCl₂ is formed.
First, we need to calculate the number of moles of CuCl₂ used:
Molar mass of CuCl₂ = atomic mass of Cu + 2 * atomic mass of Cl = 63.55 + (2 * 35.45) = 134.45 g/mol
Number of moles = mass / molar mass = 16.06 g / 134.45 g/mol ≈ 0.1194 mol
Since the reaction is 1:1, the maximum amount of PbCl₂ precipitate formed is also 0.1194 mol.
To convert to grams, we need to multiply by the molar mass of PbCl₂:
Molar mass of PbCl₂ = atomic mass of Pb + 2 * atomic mass of Cl = 207.2 + (2 * 35.45) = 278.1 g/mol
Mass of PbCl₂ precipitate = moles of PbCl₂ * molar mass of PbCl₂
= 0.1194 mol * 278.1 g/mol ≈ 33.2 g
Therefore, the maximum amount of precipitate that could be formed in this reaction is approximately 33.2 grams.