Answer:
Pb(NO₃)₂ (aq) + 2NaCl (aq) → PbCl₂ (s) ↓ + 2NaNO₃ (aq)
Explanation:
The reactants are:
Lead(II) nitrate → Pb(NO₃)₂ (aq)
Sodium chloride → NaCl (aq)
The products are:
Lead(II) chloride → PbCl₂ (s)
Sodium nitrate → NaNO₃ (aq)
Salts form nitrate are soluble. The chloride makes a precipitate with the Pb²⁺. The chemical equation for this reaction is:
Pb(NO₃)₂ (aq) + 2NaCl (aq) → PbCl₂ (s) ↓ + 2NaNO₃ (aq)
Answer:
Products are stearate anion and water.
Explanation:
Stearic acid is a 18-carbon chain molecule containing -COOH group. IUPAC name of stearic acid is octadecanoic acid.
Molecular formula of stearic acid is
.
When
is added into stearic acid,
removes a proton (
) from acidic -COOH group and forms stearate anion and water as products.
The balanced acid-base reaction is given as:

Structure of products are given below.
Lead(II) nitrate will react with iron(III) chloride to produce the precipitate lead(II) chloride as shown in the balanced reaction
2FeCl3(aq) + 3Pb(NO3)2(aq) → 2Fe(NO3)3(aq) + 3PbCl2(s)
Calculating the amount of the precipitate lead(II) chloride each reactant will produce:
mol PbCl2 = 0.050L Pb(NO3)2 (0.100mol/1L)(3mol PbCl2/3mol Pb(NO3)2)
= 0.00500mol PbCl2
mol PbCl2 = 0.050L FeCl3 (0.100mol FeCl3/1L)(3mol PbCl2/2mol FeCl3) = 0.00750mol PbCl2
The reactant Pb(NO3)2 produces a lesser amount of the precipitate PbCl2, therefore, the lead(II) nitrate is the limiting reagent for this reaction.
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