Question

How many milligrams of sodium sulfide are needed to completely react with 25.00 mL of a 0.0100 M aqueous solution of cadmium nitrate, to form a precipitate of CdS(s)?

Answer 1

Solution :-

Balanced reaction equation

Cd(NO3)2 + Na2S ------ > CdS + 2 NaNO3

Using the mole ratio of the both reactant we can calculate the amount of Na2S needed to react with 25.00 ml of 0.0100 M Cd(NO3)2

Lets first calculate the moles of Cd(NO3)2

Moles = molarity * volume in liter

Moles of Cd(NO3)2 = 0.0100 mol per L * 0.025 L = 0.00025 mol Cd(NO3)2

Mole ratio is 1 : 1 so the moles of Na2S needed are same as moles of Cd(NO3)2

Soi moles of Na2S needed = 0.00025 moles

Now lets calculate its mass in milligram

(0.00025 mol Na2S * 78.0452 g / 1 mol )*(1000 mg / 1 g ) = 19.51 mg Na2S

So the mass of Na2S needed is 19.51 mg Na2S