A solution contains 0.55 g of dissolved Sn

4+ ions. How many grams of Na2CO3 must be
added to the solution to completely precipitate all of the dissolved Sn4+?

Here is a worked example of a stoichiometry problem.

http://www.jiskha.com/science/chemistry/stoichiometry.html

To determine the amount of Na2CO3 needed to completely precipitate all of the dissolved Sn4+ ions, we need to use stoichiometry and the concept of limiting reactants.

1. Start by writing and balancing the equation for the reaction between Sn4+ ions and Na2CO3:
Sn4+ (aq) + 2 CO3^2- (aq) → SnCO3 (s)

2. Determine the molar mass of SnCO3:
Sn: 118.71 g/mol
C: 12.01 g/mol
O: 16.00 g/mol

SnCO3: 118.71 + 12.01 + (3 × 16.00) = 164.72 g/mol

3. Calculate the number of moles of Sn4+ ions in the solution:
Given mass of Sn4+ ions = 0.55 g
Molar mass of Sn4+ = 118.71 g/mol

Moles of Sn4+ = Given mass / Molar mass = 0.55 g / 118.71 g/mol

4. Use the stoichiometry of the reaction to determine the moles of SnCO3 formed. According to the balanced equation, 1 mole of Sn4+ reacts with 1 mole of SnCO3:
Moles of SnCO3 formed = Moles of Sn4+ in solution

5. Based on the stoichiometry, we can convert the moles of SnCO3 to grams:
Grams of SnCO3 = Moles of SnCO3 × Molar mass of SnCO3

Now, let's plug in the values to calculate the grams of Na2CO3 required.

6. Determine the molar mass of Na2CO3:
Na: 22.99 g/mol
C: 12.01 g/mol
O: 16.00 g/mol

Na2CO3: 2 × 22.99 + 12.01 + (3 × 16.00) = 105.99 g/mol

7. Calculate the moles of Na2CO3 required:
Moles of Na2CO3 = Grams of SnCO3 / Molar mass of SnCO3

8. Convert the moles of Na2CO3 to grams:
Grams of Na2CO3 = Moles of Na2CO3 × Molar mass of Na2CO3

By following these steps, you can calculate the number of grams of Na2CO3 needed to completely precipitate all of the dissolved Sn4+ ions in the solution.