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%
) Since there are 2 hydrogen atoms in a hydrogen molecule (
) then each carbon atom will react with 2 hydrogen molecules. Since there are 22 hydrogen molecules half as many methane molecules will be formed, so 11. Then we can subtract 11 from 34 to find the number of carbon atoms left over.