Question

0.02 M $K_{2}CR_{2} } O_{7}$ is treated with 0.288 gram of Ferrous oxalate .How much volume (in ml) of $K_{2} CR_{2} O_{7}$ is required?

Answer 1

Answer:

V = 33.33 mL

Explanation:

Here you need to equate the no. of equivalents of K2Cr2O7 to that of FeC2O4.

K2CR2O7 will change to Cr(+3), so the electron change will be 6. Thus n-factor(n¹) = 6

C204(2-) will oxidise to CO2 and the electron change will be 2. Thus n-factor(n²) = 2

No. of moles of FeC2O4 = 0.288/144 = 0.002 mol

Therefore, no. of equivalents = 2×0.002 eq

No. of moles of K2CR2O7 = 0.02 mol/L = 6×0.02 eq/L

Thus, no. of equivalents = 6×0.02×V eq

Now, 6×0.02×V = 2×0.002

V = 1/30 L = 1000/30 mL = 33.33 mL

So that will be the required volume to oxidise 0.288g of FeC2O4 completely by K2Cr2O7

Answer 2

EXPLANATION.

$\sf \to \: 0.02 \: m \: of \: k_{2} cr_{2}o_{7} \: is \: treated \: with \: 0.288 \: gram \: of \: fec_{2} o_{4} \\ \\ \sf \to \: as \: we \: know \: that \: \\ \\ \sf \to \: no \: of \: moles \: = \frac{mass}{molar \: mass} \\ \\ \sf \to \: no \: of \: moles \: of \: k_{2} cr_{2}o_{7} = 0.02m \\ \\ \sf \to \: no \: of \: moles \: of \: fec_{2} o_{4} = \frac{0.288}{144} \: m$

$\sf \to \: k_{2}cr_{2} {}^{ + 6}o_{7} \: \: + \: fe {}^{ + 2} c_{2} {}^{ + 3} o_{4} \: \longrightarrow \: fe {}^{ + 3} + 2 co_{2} {}^{ + 4} + 2cr {}^{ + 3} \\ \\ \sf \to \: n \: factor \: of \: k_{2}cr_{2}o_{7} = 6 \\ \\ \sf \to \: n \: factor \: of \: fec_{2}o_{4} \: = 3$

$\sf \to \: milli \: equivalent \: of \: k_{2}cr_{2}o_{7} \: = milli \: equivalent \: of \: fec_{2}o_{4} \\ \\ \sf \to \: M _{1} \times n_{1} \times v_{1} = M_{2} \times n_{2} \times v_{2} \\ \\ \sf \to \: 0.02 \times 6 \times v_{1} = \frac{0.288}{144} \times 3 \times 1000 \\ \\ \sf \to \: \frac{2}{100} \times 6 \times v_{1} = \frac{288}{144 \times 1000} \times 3 \times 1000 \\ \\ \sf \to \: \frac{12}{100} \times v_{1} = 6 \\ \\ \sf \to \: 2 v_{1} = 100 \implies \: v_{1} = 50ml$