Question

Write the state of matter for every substance in the chemical reaction.

Explain your limiting reactant determinations.

Mark every measured number showing the significant figures (SF).

Do not round any intermediate values. (you can show two digits past the marked SF)

Show subtraction and addition in the vertical format.

Show your final answer before rounding and then in the final form, rounded correctly.

Every number must have units and substances. You can use tables or other means to increase your efficiency, but the reader must be able to clearly identify the units and substances related to every number.

Include descriptive statements indicating what you are doing. For example: "Calculate the moles of the reactants"

EXERCISE 3.6.2: LANTHANUM OXALATE

Show the steps to answer what mass of solid lanthanum(III) oxalate nonahydrate [La₂(C₂O₄)₃•9H₂O] can be obtained from 650 mL of a 0.0170 M aqueous solution of LaCl₃ by adding a stoichiometric amount of sodium oxalate?

Strategy:

1. Check the chemical equation to make sure it is balanced as written; balance if necessary. Then calculate the number of moles of [Au(CN)₂]⁻ present by multiplying the volume of the solution by its concentration.
2. From the balanced chemical equation, use a mole ratio to calculate the number of moles of gold that can be obtained from the reaction. To calculate the mass of gold recovered, multiply the number of moles of gold by its molar mass.

Answer

3.89 g

Answer 1

Given molarity of LaL₃ solution, C = 0.0170 mol/L

Volume of  LaL₃ solution taken, V = 650 mL * (1L / 1000 mL) = 0.650 L

=> Moles of LaL₃ initially taken = C*V = 0.0170 mol/L * 0.650 L = 0.01105 mol

The balanced equation for reaction of  LaCl₃ and sodium oxalate, Na₂C₂O₄ to form solid lanthanum(III) oxalate nonahydrate, La₂(C₂O₄)₃.9H₂O(s) is:

2LaCl₃(aq) + 3Na₂C₂O₄(aq) +9H₂O(l) ----> La₂(C₂O₄)₃.9H₂O(s) + 6NaCl(aq)

The coefficient of LaCl₃(aq) in the balanced equation is 2 and that of La₂(C₂O₄)₃.9H₂O(s) is 1.

Hence mole ration between La₂(C₂O₄)₃.9H₂O(s) and LaCl₃(aq) is 1:2.

=> Moles of La₂(C₂O₄)₃.9H₂O(s) formed = 0.01105 mol LaCl₃(aq) * [1 mol La₂(C₂O₄)₃.9H₂O(s) / 2 mol LaCl₃(aq) ]

= 0.01105 * (1/2) mol  La₂(C₂O₄)₃.9H₂O(s)

= 0.005525 mol La₂(C₂O₄)₃.9H₂O(s)

Molar mass of La₂(C₂O₄)₃.9H₂O(s) = 704.0 g/mol

=> Mass of La₂(C₂O₄)₃.9H₂O(s) obtained = 0.005525 mol * (704.0 g / 1 mol) = 3.89 g (Answer)