How Much Iron Is Required to Treat Perchloroethene Contamination in Water?

[hils0005]

1. Homework Statement [/b]

Given:
C2CL4 + Fe $$\rightarrow$$ C2H4 + CL- + Fe2+

A reservoir contains 5million gallons of water which is contaminated with 4.2mg/L of Perchloroethene(C2CL4). You want too reduce it to harmless ethene(C2H4) by reacting it with Fe. because of this only about 20g of the added Fe is reactive. If you need to treat 5million gallons of water, how many grams of Fe will you need to treat the Perchloroethene(PCE)?

Homework Equations

I first balanced the equation
C2CL4 + 4H+ + 4Fe $$\rightarrow$$ C2H4 + 4CL- + 4Fe^2+

Concentration of PCE=4.2mg/L
5mill gallons of water = 1.89 x 10^7 L

I do not know where to go from here

The Attempt at a Solution

 

[symbolipoint]

How many times bigger is the amount of PCE in the 5 million gallon tank than in the elementary reaction? How many milligrams, and therefore moles, of PCE is in the tank? I believe some information is missing related to your "20 grams" comment about the Fe.

 

[Blueshift5]

To determine the amount of Fe needed, we can use the balanced equation to calculate the moles of PCE present in 5 million gallons of water. Then, using the mole ratio from the balanced equation, we can calculate the moles of Fe needed to react with all of the PCE. Finally, we can convert the moles of Fe to grams using its molar mass.

First, we need to convert the concentration of PCE from mg/L to moles/L. This can be done by dividing by its molar mass (165.83 g/mol).

4.2 mg/L / 165.83 g/mol = 2.5 x 10^-5 mol/L

Next, we can use this concentration to calculate the moles of PCE present in 5 million gallons of water.

2.5 x 10^-5 mol/L x 1.89 x 10^7 L = 473.25 mol PCE

From the balanced equation, we can see that 4 moles of Fe are needed to react with 1 mole of PCE. Therefore, the moles of Fe needed will be:

473.25 mol PCE x (4 mol Fe / 1 mol PCE) = 1893 mol Fe

Finally, we can convert the moles of Fe to grams by multiplying by its molar mass (55.845 g/mol).

1893 mol Fe x 55.845 g/mol = 105.6 kg Fe

Therefore, to treat 5 million gallons of water contaminated with 4.2 mg/L of PCE, we will need approximately 105.6 kg (or 105,600 g) of Fe. However, since we know that only 20 g of Fe will be reactive, we will need to add more Fe in order to ensure that all of the PCE is treated.