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

Na₂S(aq) + Cd(NO₃)₂(aq) = CdS(s) + 2NaNO₃(aq)

v=25.00 mL
c=0.0100 mmol/mL
M(Na₂S)=78.046 mg/mmol

n(Na₂S)=n{Cd(NO₃)₂}=cv

m(Na₂S)=M(Na₂S)n(Na₂S)=M(Na₂S)cv

m(Na₂S)=78.046*0.0100*25.00≈19.5 mg

Answer 2

Answer: The mass of sodium sulfide needed is 195000 mg.

Explanation:

To calculate the moles of cadmium nitrate, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$

Molarity of cadmium nitrate = 0.0100 M

Volume of cadmium nitrate = 25.0 mL = 0.025 L   (Conversion factor: 1 L = 1000 mL)

Putting values in above equation, we get:

$0.0100mol/L=\frac{\text{Moles of cadmium nitrate}}{0.025L}\\\\\text{Moles of cadmium nitrate}=2.5\times 10^{-4}mol$

The chemical reaction of cadmium nitrate with sodium sulfide, the equation follows:

$Cd(NO_3)_2+Na_2S\rightarrow CdS+2NaNO_3$

By Stoichiometry of the reaction:

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

So, $2.5\times 10^{-4}mol$ of cadmium nitrate will react with = $\frac{1}{1}\times 2.5\times 10^{-4}=2.5\times 10^{-4}mol$ of sodium sulfide.

• To calculate the mass of sodium sulfide, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Moles of sodium sulfide = $2.5\times 10^{-4}mol$

Molar mass of sodium sulfide = 78 g/mol

Putting values in above equation, we get:

$2.5\times 10^{-4}mol=\frac{\text{Mass of sodium sulfide}}{78g/mol}\\\\\text{Mass of sodium sulfide}=195g$

Now, converting this mass into milligrams, we use the conversion factor:

1 g = 1000 mg

So, $195g=195\times 1000=195000mg$

Hence, the mass of sodium sulfide needed is 195000 mg.