In a simulation mercury removal from industrial wastewater, 0.020 L of 0.10 M sodium sulfide reacts with 0.050 L of 0.010 M mercury(II) nitrate. How many grams of mercury(II) sulfide (232.2 g/mol) form

 0.1161 grams of mercury(II) sulfide) form.

Explanation:

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

    .....(1)

a) Molarity of solution = 0.10 M

Volume of solution = 0.020 L

Putting values in equation 1, we get:

b) Molarity of solution = 0.010 M

Volume of solution = 0.050 L

Putting values in equation 1, we get:

According to stoichiometry :

1 mole of reacts with 1 mole of

Thus 0.0005 moles of reacts with= moles of

Thus is the limiting reagent and is the excess reagent.

According to stoichiometry :

1 mole of forms=  1 mole of

Thus 0.0005 moles of forms= moles of

mass of

Thus 0.1161 grams of mercury(II) sulfide) form.

3 1/4 %

explanation:

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