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Answer :
Final Answer:
a. Balanced Equation: [tex]\(2AgNO_2 + Na_2S \rightarrow Ag_2S + 2NaNO_2\)[/tex]
b. 71 grams of silver sulfide will be produced.
c. 0 grams of silver nitrite will remain at the end of the reaction.
Explanation:
a. To balance the given chemical equation, we first write down the unbalanced equation:
[tex]\(AgNO_2 + Na_2S \rightarrow Ag_2S + NaNO_2\)\\[/tex]
Balancing the equation, we get
[tex]\(2AgNO_2 + Na_2S \rightarrow Ag_2S + 2NaNO_2\)[/tex]
This balanced equation shows the correct stoichiometry of the reaction, indicating that 2 moles of silver nitrite react with 1 mole of sodium sulfide to produce 1 mole of silver sulfide and 2 moles of sodium nitrite.
b. To calculate the number of grams of silver sulfide produced, we need to find the molar mass of [tex]\(Ag_2S\)[/tex] which is
[tex]\(2 \times 107.87 \, \text{g/mol (Ag)} + 1 \times 32.07 \, \text{g/mol (S)} = 243.81 \, \text{g/mol}\)[/tex]
Using the balanced equation, we can see that 2 moles of [tex]\(Ag_2S\)[/tex]are produced from 2 moles of [tex]\(AgNO_2\)[/tex]. Since the total mass of silver nitrite is 71 grams [tex](\(2 \times 35.5 \, \text{g}\))[/tex], the mass of silver sulfide produced will also be 71 grams.
c. According to the balanced equation, 2 moles of [tex]\(AgNO_2\)[/tex]react with 1 mole of [tex]\(Na_2S\)[/tex]. Since both substances are provided in equal masses (35.5 grams each), all the silver nitrite will react completely, leaving 0 grams of silver nitrite at the end of the reaction.
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