Question

what volume of 0.200 m k2c2o4 is required to react completely with 30.0 ml of 0.100 m fe(no3)3? 2fe(no3)3 3k2c2o4fe2(c2o4)3 6kno3

Answer 1

To react completely with 30.0 mL of 0.100 M Fe(NO3)3, you will need 22.5 mL of 0.200 M K2C2O4.

Step-by-step explanation:

To determine the volume of 0.200 M K2C2O4 required to react completely with 30.0 mL of 0.100 M Fe(NO3)3, we need to use the stoichiometry of the balanced chemical equation. The balanced equation shows that 2 moles of Fe(NO3)3 react with 3 moles of K2C2O4 to form 1 mole of Fe2(C2O4)3 and 6 moles of KNO3.

First, let's calculate the moles of Fe(NO3)3:

moles of Fe(NO3)3 = (0.100 M) * (0.030 L) = 0.003 mol

According to the stoichiometry of the equation, 2 moles of Fe(NO3)3 react with 3 moles of K2C2O4. So, the moles of K2C2O4 required would be:

moles of K2C2O4 = (0.003 mol Fe(NO3)3) * (3 mol K2C2O4 / 2 mol Fe(NO3)3) = 0.0045 mol

To calculate the required volume, we can use the molarity equation:

Molarity = moles / volume

0.200 M = 0.0045 mol / volume

Solving for volume:

volume = 0.0045 mol / 0.200 M = 0.0225 L = 22.5 mL

Therefore, 22.5 mL of 0.200 M K2C2O4 is required to react completely with 30.0 mL of 0.100 M Fe(NO3)3.

Answer 2

To react completely with 30.0 mL of 0.100 M Fe(NO₃)₃, 22.5 mL of 0.200 M K₂C₂O₄ is necessary, based on the stoichiometry of the reaction equation.

Step-by-step explanation:

To calculate the volume of 0.200 M K₂C₂O₄ needed to react completely with 30.0 mL of 0.100 M Fe(NO₃)₃, we must first understand the stoichiometry of the reaction. According to the balanced equation:

2Fe(NO₃)₃(aq) + 3K₂C₂O₄(aq) → Fe₂(C₂O₄)₃(s) + 6KNO₃(aq)

It shows that 2 moles of Fe(NO₃)₃ react with 3 moles of K₂C₂O₄. Using this information:

• First, calculate the moles of Fe(NO₃)₃: Moles = Molarity × Volume = 0.100 mol/L × 0.0300 L = 0.0030 mol.
• From the stoichiometry, for every 2 moles of Fe(NO₃)₃, 3 moles of K₂C₂O₄ are required. Therefore, for 0.0030 mol of Fe(NO₃)₃, we need (3/2) × 0.0030 mol = 0.0045 mol of K₂C₂O₄.
• Finally, calculate the volume of 0.200 M K₂C₂O₄ needed: Volume = Moles / Molarity = 0.0045 mol / 0.200 mol/L = 0.0225 L or 22.5 mL.

Therefore, 22.5 mL of 0.200 M K₂C₂O₄ is required to react completely with 30.0 mL of 0.100 M Fe(NO₃)₃.