Final answer:
To determine the amount of precipitate formed when iron(III) chlorate reacts with sodium hydroxide, the moles of iron(III) chlorate must be calculated using its mass and molar mass, followed by a stoichiometric conversion to find the precipitate's mass.
Step-by-step explanation:
The student is tasked with determining the amount of precipitate formed when 110.00 grams of iron(III) chlorate is reacted with excess sodium hydroxide. To solve this problem, we must first write the balanced chemical equation for the reaction, then calculate the number of moles of iron(III) chlorate, followed by using stoichiometry to find the amount of precipitate formed.
Assuming the reaction is:
Fe(ClO3)3(aq) + 3NaOH(aq) → Fe(OH)3(s) + 3NaClO3(aq)
This reaction suggests that one mole of iron(III) chlorate reacts with three moles of sodium hydroxide to form one mole of iron(III) hydroxide as precipitate and three moles of sodium chlorate. Here's how you can calculate the precipitate:
- Determine the molar mass of iron(III) chlorate.
- Calculate the number of moles of iron(III) chlorate in 110.00 grams.
- Use the mole ratio from the balanced equation to find the moles of iron(III) hydroxide precipitate.
- Calculate the mass of precipitate using the molar mass of iron(III) hydroxide.
To determine if other solid products form from reactions with solute ions, solubility rules and reaction stoichiometry are applied. For example, mixing cesium bromide and lead (II) nitrate will produce lead (II) bromide precipitate as these two salts react to form an insoluble product. Similarly, the reaction between silver nitrate and sodium chloride yields a precipitate of silver chloride. The solubility of potential precipitates such as barium sulfate can be predicted by understanding solubility rules and calculating the reaction quotient Q in comparison to the solubility product constant Ksp.