If a silver nitrate solution is added to excess sodium sulfide, this reaction takes place:
[tex]2 AgNO_3(a q)+Na_2 S(a q) \rightarrow Ag_2 S(s)+2 NaNO_3(a q)[/tex]
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. [tex]1.49 g[/tex]
B. 4.55 g
C. 9.30 g
Answer (2)
Calculate the moles of A g N O 3 : m o l es = 2.50 × 0.0150 = 0.0375 \t e x t m o l es .
Determine the moles of A g 2 S : m o l e s A g 2 S = \t f r a c 1 2 × 0.0375 = 0.01875 \t e x t m o l es .
Calculate the molar mass of A g 2 S : M o l a r M a s s A g 2 S = 2 × 107.87 + 32.07 = 247.81 \t e x t g / m o l .
Calculate the mass of A g 2 S : ma s s A g 2 S = 0.01875 × 247.81 = 4.646 \t e x t g .
The mass of silver sulfide produced is 4.646 g .
Explanation
Understanding the Problem First, we need to understand the problem. We are given a reaction between silver nitrate ( A g N O 3 ) and sodium sulfide ( N a 2 S ) that produces silver sulfide ( A g 2 S ) and sodium nitrate ( N a N O 3 ). We are given the volume and molarity of the silver nitrate solution and asked to find the mass of silver sulfide produced, assuming the reaction goes to completion.
Calculate Moles of Silver Nitrate Next, we need to calculate the moles of A g N O 3 used in the reaction. We can use the formula: m o l es = M o l a r i t y × V o l u m e Given that the molarity of A g N O 3 is 2.50 M and the volume is 0.0150 L, we have: m o l e s A g N O 3 = 2.50 × 0.0150 = 0.0375 \t e x t m o l es
Determine Moles of Silver Sulfide Now, we need to determine the moles of A g 2 S produced. From the balanced chemical equation: 2 A g N O 3 ( a q ) + N a 2 S ( a q ) \t o A g 2 S ( s ) + 2 N a N O 3 ( a q ) We see that 2 moles of A g N O 3 produce 1 mole of A g 2 S . Therefore, the moles of A g 2 S produced is half the moles of A g N O 3 used: m o l e s A g 2 S = \t f r a c 1 2 × m o l e s A g N O 3 = \t f r a c 1 2 × 0.0375 = 0.01875 \t e x t m o l es
Calculate Molar Mass of Silver Sulfide Next, we calculate the molar mass of A g 2 S . Using the periodic table, the molar mass of silver (Ag) is approximately 107.87 g/mol, and the molar mass of sulfur (S) is approximately 32.07 g/mol. Therefore, the molar mass of A g 2 S is: M o l a r M a s s A g 2 S = 2 × M o l a r M a s s A g + M o l a r M a s s S = 2 × 107.87 + 32.07 = 215.74 + 32.07 = 247.81 \t e x t g / m o l
Calculate Mass of Silver Sulfide Finally, we calculate the mass of A g 2 S produced using the formula: ma ss = m o l es × M o l a r M a ss ma s s A g 2 S = m o l e s A g 2 S × M o l a r M a s s A g 2 S = 0.01875 × 247.81 = 4.646 \t e x t g Therefore, the mass of silver sulfide produced is approximately 4.646 g.
Final Answer The mass of silver sulfide ( A g 2 S ) produced is approximately 4.646 g.
Examples
This type of calculation is essential in environmental chemistry when assessing the impact of heavy metals on water quality. For instance, if silver ions are released into a wastewater stream containing sulfide, they can precipitate as silver sulfide. Knowing the concentration of silver ions and sulfide allows us to calculate the amount of silver sulfide that will form, which helps in designing effective treatment processes to remove these pollutants and protect aquatic ecosystems. This ensures compliance with environmental regulations and safeguards public health by preventing the accumulation of toxic substances in the environment.
When 0.0150 L of a 2.50 M silver nitrate solution is reacted with excess sodium sulfide, approximately 4.646 g of silver sulfide ( A g 2 S ) is produced. This aligns with option B, which states 4.55 g as the closest answer. Therefore, option B is selected.
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