explain how to set up: what mass of iron(III) carbonate would be produced if 545 ml of 0.150 M iron(III) choloride were added to a solution containing excess sodium carbonate.
Given reaction of 2FECL3(aq) + 3Na2CO3(aq)---- Fe2(CO3)3 (s) + 6 NaCl (aq)
thank you
Here is a detailed explanation of a stoichiometry problem.
http://www.jiskha.com/science/chemistry/stoichiometry.html
To determine the mass of iron(III) carbonate produced, you need to use the given information and the balanced chemical equation.
Step 1: Write the balanced chemical equation.
The balanced chemical equation is:
2FeCl3(aq) + 3Na2CO3(aq) → Fe2(CO3)3(s) + 6NaCl(aq)
Step 2: Determine the limiting reactant.
In this case, the excess sodium carbonate indicates that iron(III) chloride is the limiting reactant. To find the amount of iron(III) chloride used, you need to calculate the moles using the given concentration and volume.
Molarity (M) = moles/volume (in L)
Given: Concentration of iron(III) chloride (FeCl3) = 0.150 M, Volume of iron(III) chloride (FeCl3) = 545 mL = 0.545 L
Moles of FeCl3 = 0.150 M × 0.545 L.
Step 3: Use the stoichiometry of the balanced equation to convert moles of FeCl3 to moles of Fe2(CO3)3.
From the balanced equation, 2 moles of FeCl3 produce 1 mole of Fe2(CO3)3.
Step 4: Convert moles of Fe2(CO3)3 to grams.
To convert moles of Fe2(CO3)3 to grams, you need to multiply the moles by the molar mass of Fe2(CO3)3.
The molar mass of Fe2(CO3)3 = 2(55.85 g/mol) + 3(12.01 g/mol + 16.00 g/mol) × 3.
Step 5: Calculate the mass of Fe2(CO3)3.
Using the obtained moles and molar mass, calculate the mass of Fe2(CO3)3.
Mass of Fe2(CO3)3 = Moles of Fe2(CO3)3 × Molar mass of Fe2(CO3)3.
Now you have the formula to calculate the mass of iron(III) carbonate produced when 545 mL of 0.150 M iron(III) chloride is added to a solution containing excess sodium carbonate.