Heavy metal ions like lead(ii) can be precipitated from laboratory wastewater by adding sodium sulfide, na2s. will all the lead be removed from 11.2 ml of 7.10×10-3 m pb(no3)2 upon addition of 12.4 ml of 0.0117 m na2s? if all the lead is removed, how many moles of lead is this? if not, how many moles of pb remain?

The moles of lead removed is

Explanation:

To calculate the number of moles for given molarity, we use the equation:

Molarity of sodium sulfide solution = 0.0117 M

Volume of solution = 12.4 mL = 0.0124 L    (Conversion factor:  1 L = 1000 mL)

Putting values in equation 1, we get:

Molarity of lead nitrate solution = M

Volume of solution = 11.2 mL = 0.0112 L

Putting values in equation 1, we get:

The chemical equation for the reaction of sodium sulfide and lead nitrate follows:

By Stoichiometry of the reaction:

1 mole of lead nitrate reacts with 1 mole of sodium sulfide.

So, will react with = of sodium sulfide.

As, given amount of sodium sulfide is more than the required amount. So, it is considered as an excess reagent.

Thus, lead nitrate is considered as a limiting reagent because it limits the formation of product.

As, all the lead from lead nitrate is getting converted to lead sulfide. So, all the lead (II) is removed from the solution.

By Stoichiometry of the reaction:

1 mole of lead nitrate produces 1 mole of lead sulfide.

So, will produce = of lead sulfide.

Hence, the moles of lead removed is

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