A 2.80 mol/L solution of iron (II) dichromate is combined with 5.00L of a
2.25 mol/L acidic solution of tin (II) bromide. Calculate the volume of iron (II) dichromate required.
FeCr2O7 + SnBr2 --> SnCr2O7 + FeBr2
Is this the right formula? or no?
To determine the volume of iron (II) dichromate required, we need to use stoichiometry and balance the chemical equation first.
The balanced equation for the reaction is:
3 FeCr2O7 + 16 SnBr2 → 8 SnCr2O7 + 6 FeBr2
This means that 3 moles of FeCr2O7 react with 16 moles of SnBr2 to produce 8 moles of SnCr2O7 and 6 moles of FeBr2.
Next, we can use the stoichiometry to calculate the volume of iron (II) dichromate required.
Step 1: Calculate the number of moles of SnBr2 in the 5.00L solution:
moles of SnBr2 = concentration of SnBr2 × volume of SnBr2 solution = 2.25 mol/L × 5.00 L = 11.25 mol
Step 2: Determine the required amount of FeCr2O7 based on the balanced equation:
moles of FeCr2O7 = (moles of SnBr2 / 16) × 3 = (11.25 mol / 16) × 3 = 2.109 moles
Step 3: Convert the moles of FeCr2O7 to volume using its concentration:
volume of FeCr2O7 = moles of FeCr2O7 / concentration of FeCr2O7
volume of FeCr2O7 = 2.109 mol / 2.80 mol/L ≈ 0.753 L
Therefore, the volume of iron (II) dichromate required is approximately 0.753 liters.