If it takes 35.0 ml of 0.250 m fe2+ to titrate 100 ml of a solution containing cr2o72-, what is the molar concentration of cr2o72-?

0.0146 M

Explanation:

The relevant chemical reaction equation is as follows:

Cr₂O₇²⁻ + 6Fe²⁺ + 14H⁺ ⇒ 2Cr³⁺ +6Fe³⁺ + 7H₂O

In order to find the concentration of Cr₂O₇²⁻, the number of moles of Cr₂O₇²⁻ in the solution must be found. This can be found by relating it to the number of moles of the analyte (Fe²⁺), which can be calculated as follows:

(35.0 mL)(0.250mol/L) = 8.75 mmol Fe²⁺

This is the amount of Fe²⁺ required to react with all of the Cr₂O₇²⁻ present.The moles of Fe²⁺ and Cr₂O₇²⁻ in the reaction are related as follows:

8.75 mmol Fe²⁺ x (Cr₂O₇²⁻ / 6Fe²⁺) = 1.4583...mmol Cr₂O₇²⁻

The concentration of the Cr₂O₇²⁻ solution is then calculated as follows:

(1.4583...mmol) / 100mL = 0.146 mol/L = 0.146 M

n2(g) + 3h2-> 2nh3(g)   this is the balanced equation

note the mole ratio between n2, h2 and nh3.   it is 1 : 3 : 2   this will be important.

moles n2 present = 28.0 g n2 x 1 mole n2/28 g = 1 mole n2 present

moles h2 present = 25.0 g h2 x 1 mole h2/2 g = 12.5 moles h2 present

based on mole ratio, n2 is limiting in this situation because there is more than enough h2 but not enough n2.

moles nh3 that can be produced = 1 mole n2 x 2 moles nh3/mole n2 = 2 moles nh3 can be produced

grams of nh3 that can be produced = 2 moles nh3 x 17 g/mole = 34 grams of nh3 can be produced

note:   the key to this problem is recognizing that n2 is limiting, and therefore limits how much nh3 can be produced.