Suppose an epa chemist tests a 200.ml sample of groundwater known to be contaminated with iron(ii) chloride, which would react with silver nitrate solution like this: fecl2 (aq) + 2agno3 (aq) → 2agcl (s) + feno32 (aq) the chemist adds 14.0m m silver nitrate solution to the sample until silver chloride stops forming. she then washes, dries, and weighs the precipitate. she finds she has collected 6.9mg of silver chloride. calculate the concentration of iron(ii) chloride contaminant in the original groundwater sample. be sure your answer has the correct number of significant digits.

The concentration of iron(II) chloride contaminant in the original groundwater sample is .

Explanation:

Amount of silver chloride recovered = 6.9 mg = 0.0069 g[/tex]

Moles of silver chloride =

According to reaction. 2 moles of silver chloride are obtained from 1 mole of iron(II) chloride.

Then moles of silver chloride will be obtained from:

Moles of iron(II) chloride =

Volume of a sample of ground water = 200 mL = 0.2 L

The concentration of iron(II) chloride contaminant in the original groundwater sample is .

the equation that correctly represent the reaction for formation of ammonia is

n2+ 3h2 → 2nh3 (answer d)

1 mole of nitrogen gas (n2) react with 3 moles of hydrogen gas (h2) to form 2 moles of ammonia ( nh3). this is in a process known as haber process were iron is used as a catalyst and reaction take place that a higher temperature and pressure. the process is exothermic hence energy is released.

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.