Answer:
a) Pb(NO₃)₂(aq) + Na₂Cr₂O₄(aq) ⇄ 2NaNO₃(aq) + Pb(Cr₂O₄)(s)
b) 67.6%
Explanation:
a) Nitrate is the ion NO₃⁻, and lead(II) forms the ion Pb⁺², so the compound lead(III) nitrate is Pb(NO₃)₂. (First, the cation, then the anion, with charges replaced without the signal).
Chromate is the ion Cr₂O₄⁻² and sodium forms the ion Na⁺, so the sodium chromate is Na₂Cr₂O₄. Both of them are in solutions, so they will be in an aqueous state.
In the reaction, the anions and cations will replace and will form: NaNO₃ and Pb(Cr₂O₄). The nitrates formed by metals from group 1, such as sodium, are soluble, so it will not forme a precipitated. So, the precipitated is PbCr₂O₄, and the balanced reaction is:
Pb(NO₃)₂(aq) + Na₂Cr₂O₄(aq) ⇄ 2NaNO₃(aq) + Pb(Cr₂O₄)(s)
b) The molar masses are: Pb(NO₃)₃ = 331,2 g/mol; Na₂Cr₂O₄ = 162 g/mol; Pb(Cr₂O₄) = 323,2 g/mol.
First, let's find what is the limiting reactant, doing the stoichiometry calculus for the reactants. Let's suppose that Na₂Cr₂O₄ is the limiting so:
1 mol of Pb(NO₃)₂ ------------------------------ 1 mol of Na₂Cr₂O₄
Transforming to mass (mass = moles * molar mass):
331,2 g of Pb(NO₃)₂ ------------------------- 162 g/mol of Na₂Cr₂O₄
x ------------------------- 12.38
By a simple direct three rule:
162x = 4100.256
x = 25.3 g of Pb(NO₃)₂
This is higher than what is put in the reaction, so Pb(NO₃)₂ is the limiting reactant, and Na₂Cr₂O₄ is in excess. So, let's do the stoichiometric calculus for the limiting reactant and the solid formed:
1 mol of Pb(NO₃)₂ ----------------------- 1 mol of Pb(Cr₂O₄)
Transforming to mass:
331.2 g of Pb(NO₃)₂ ------------------- 323.2 g of Pb(Cr₂O₄)
11.39 g ------------------- y
By a simple direct three rule:
331.2 y = 3681.248
y = 11.115 g
The yield is the mass formed divided by the stoichiometric result multiplied by 100%:
yield = (7.52/11.115)*100% = 67.6%
