Iron(II) can be oxidized by an acidic K2Cr2O7 solution according to the net ionic equation.

Cr2O72− + 6 Fe2+ + 14 H+ 2 Cr3+ + 6 Fe3+ + 7 H2O
If it takes 44.0 mL of 0.0250 M K2Cr2O7 to titrate 25.0 mL of a solution containing Fe2+, what is the molar concentration of Fe2+?

0.156 M Fe2+

How many moles of molybdenum are in 1.235 g

To find the molar concentration of Fe2+, we can use the stoichiometry of the balanced equation and the given information about the titration.

First, let's convert the volume of K2Cr2O7 solution used in the titration to moles of K2Cr2O7:

0.044 L * 0.0250 mol/L = 0.0011 mol K2Cr2O7

According to the balanced equation, the stoichiometric ratio between K2Cr2O7 and Fe2+ is 1:6. Therefore, the moles of Fe2+ in the solution titrated can be calculated as:

0.0011 mol K2Cr2O7 * (6 mol Fe2+ / 1 mol K2Cr2O7) = 0.0066 mol Fe2+

Next, we calculate the moles of Fe2+ in the 25.0 mL of the Fe2+ solution:

25.0 mL * (1 L / 1000 mL) * 0.0066 mol Fe2+ / 0.025 L = 0.0165 mol Fe2+

Finally, we calculate the molar concentration of Fe2+:

Molar concentration of Fe2+ = Moles of Fe2+ / Volume of Fe2+ solution (in L)
Molar concentration of Fe2+ = 0.0165 mol / 0.025 L = 0.66 mol/L

Thus, the molar concentration of Fe2+ in the solution is 0.66 mol/L.

moles Cr2O7^-2 = M x L = ??

Using the coefficients in the balanced equation, convert moles Cr2O7^-2 to moles Fe(II).
Then M Fe(II) = moles Fe(II)/L Fe(II)