The dichromate ion oxidises stannous ions in solution according to the reaction : 3Sn+14H+Cr2O7---->3Sn+2Cr+7H2O... How many ml of K2Cr2O7 solution of 0.5M with 0.5 mole of tin(ll) chloride in solution.

To determine the volume of potassium dichromate (K2Cr2O7) solution needed to react with 0.5 moles of stannous chloride (SnCl2), we first need to analyze the balanced chemical reaction you provided. The reaction shows that dichromate ions (Cr2O7^2-) oxidize stannous ions (Sn^2+) in an acidic medium. Let's break this down step by step.

Understanding the Reaction Stoichiometry

The balanced equation is:

3 Sn^2+ + Cr2O7^2- + 14 H+ → 3 Sn^4+ + 2 Cr^3+ + 7 H2O

From this equation, we can see that:

• 3 moles of Sn^2+ react with 1 mole of Cr2O7^2-.
• This means that for every mole of dichromate, 3 moles of stannous ions are required.

Calculating Moles of K2Cr2O7 Needed

Since you have 0.5 moles of SnCl2, we can calculate the moles of K2Cr2O7 needed:

Using the stoichiometric ratio:

1 mole of Cr2O7^2- reacts with 3 moles of Sn^2+

Therefore, the moles of Cr2O7^2- required for 0.5 moles of Sn^2+ is:

0.5 moles Sn^2+ × (1 mole Cr2O7^2- / 3 moles Sn^2+) = 0.1667 moles of Cr2O7^2-.

Finding the Volume of K2Cr2O7 Solution

Now that we know we need 0.1667 moles of K2Cr2O7, we can use the molarity of the K2Cr2O7 solution to find the volume required. The molarity (M) is defined as:

M = moles of solute / volume of solution in liters

Rearranging this formula gives us:

Volume (L) = moles of solute / M

Substituting the values we have:

Volume (L) = 0.1667 moles / 0.5 M = 0.3334 L.

Converting Liters to Milliliters

Since we need the volume in milliliters, we convert liters to milliliters:

0.3334 L × 1000 mL/L = 333.4 mL.

Final Answer

Thus, you would need approximately 333.4 mL of a 0.5 M K2Cr2O7 solution to completely react with 0.5 moles of stannous chloride in solution. This calculation demonstrates the importance of understanding stoichiometry in chemical reactions, as it allows us to predict the amounts of reactants and products involved.