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Margarita [4]
3 years ago
13

How many milligrams of sodium sulfide are needed to completely react with 25.00 ml of a 0.0100 m aqueous solution of cadmium nit

rate, to form a precipitate of cds(s)?
Chemistry
2 answers:
german3 years ago
8 0

<u>Answer:</u> The mass of sodium sulfide needed is 195000 mg.

<u>Explanation:</u>

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.

NARA [144]3 years ago
5 0
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
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