Write the net equations for this process and determine the mass of iron metal oxidized by 50.0mL of a 0.250mol/L acified K2Cr2O7(aq) solution
The net equation for the process can be represented as:
6 Fe (s) + K2Cr2O7 (aq) + 14 H+ (aq) → 6 Fe3+ (aq) + 2 Cr3+ (aq) + 2 K+ (aq) + 7 H2O (l)
To determine the mass of iron metal oxidized, we need to know the stoichiometric ratio between iron and K2Cr2O7 in the reaction.
From the balanced equation, we can see that 6 moles of Fe react with 1 mole of K2Cr2O7. This means that for every 6 moles of Fe, 1 mole of K2Cr2O7 is required.
To calculate the mass of iron, we need to convert the volume of the K2Cr2O7 solution to moles, and then use the stoichiometric ratio to determine the moles of iron reacted.
First, let's convert the volume of the solution to moles:
Volume of solution = 50.0 mL = 0.0500 L
Concentration of K2Cr2O7 solution = 0.250 mol/L
Moles of K2Cr2O7 = Concentration × Volume = 0.250 mol/L × 0.0500 L = 0.0125 mol
Using the stoichiometric ratio, we can determine the moles of Fe reacted:
Moles of Fe = (0.0125 mol K2Cr2O7) × (6 mol Fe / 1 mol K2Cr2O7) = 0.0750 mol Fe
To determine the mass of iron, we need to know the molar mass of iron (Fe):
Molar mass of Fe = 55.845 g/mol
Mass of iron = Moles of Fe × Molar mass of Fe = 0.0750 mol × 55.845 g/mol ≈ 4.19 g
Therefore, approximately 4.19 grams of iron metal will be oxidized by the 0.250 mol/L acified K2Cr2O7(aq) solution.