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Select the correct answer.
If a silver nitrate solution is added to excess sodium sulfide, this reaction takes place: 2AgNO3(aq) + Na2S(aq) → Ag2S(s) + 2NaNO3(aq).
Suppose you use 0.0150 liter of a 2.50 M solution of silver nitrate. Assuming the reaction goes to completion, how much silver sulfide is produced? Use the periodic table.
A. 1.49 g
B. 4.65 g
C. 9.30 g
D. 18.6 g

Respuesta :

Answer:- B. 4.65 g.

Solution:- The given balanced equation is:

[tex]2AgNO_3(aq)+Na_2S(aq)\rightarrow Ag_2S(s)+2NaNO_3(aq)[/tex]

It asks to calculate the mass of silver sulfide formed by when 0.0150 liters of 2.50 M of silver nitrate are used.

Moles of silver nitrate are calculated on multiplying it's liters by its molarity and then on multiplying by mol ratio, the moles of silver sulfide are calculated. These moles are multiplied by the molar mass to convert to the grams.

Molar mass of [tex]Ag_2S[/tex] = 2(107.87)+32.06  = 247.8 g per mol

The dimensional set up for the complete problem is:

[tex]0.0150L(\frac{2.50molAgNO_3}{1L})(\frac{1molAg_2S}{2molAgNO_3})(\frac{247.8gAg_2S}{1molAg_2S})[/tex]

= [tex]4.65gAg_2S[/tex]

So, the correct choice is B. 4.65 g.


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