Given the reaction: Na2S(aq) + 2 AgNO3(aq) ? Ag2S(s) + 2 NaNO3(aq). How many grams of silver(I) sulfite will form when 5.34 g of silver(I) nitrate and an excess of sodium sulfide are reacted together? (The formula weight of Na2S is 78.04 g/mol, AgNO3 is 169.87 g/mol, Ag2S is 247.80 g/mol, and NaNO3 is 84.99 g/mol.)
The correct answer and explanation is :
To solve this problem, we need to follow these steps:
- Write down the balanced chemical equation: Na2S(aq)+2AgNO3(aq)→Ag2S(s)+2NaNO3(aq)\text{Na}_2\text{S(aq)} + 2 \text{AgNO}_3\text{(aq)} \rightarrow \text{Ag}_2\text{S(s)} + 2 \text{NaNO}_3\text{(aq)} This tells us that 2 moles of silver(I) nitrate (AgNO3\text{AgNO}_3) react with 1 mole of sodium sulfide (Na2S\text{Na}_2\text{S}) to form 1 mole of silver(I) sulfide (Ag2S\text{Ag}_2\text{S}).
- Determine the molar amount of silver(I) nitrate: The given mass of silver(I) nitrate is 5.34 grams, and the molar mass of silver(I) nitrate (AgNO3\text{AgNO}_3) is 169.87 g/mol. To calculate the number of moles of AgNO3\text{AgNO}_3, use the formula: moles of AgNO3=mass of AgNO3molar mass of AgNO3\text{moles of AgNO}_3 = \frac{\text{mass of AgNO}_3}{\text{molar mass of AgNO}_3} moles of AgNO3=5.34 g169.87 g/mol=0.0315 mol\text{moles of AgNO}_3 = \frac{5.34 \, \text{g}}{169.87 \, \text{g/mol}} = 0.0315 \, \text{mol}
- Determine the amount of silver(I) sulfide formed: From the balanced equation, 2 moles of AgNO3\text{AgNO}_3 produce 1 mole of Ag2S\text{Ag}_2\text{S}. So, the number of moles of Ag2S\text{Ag}_2\text{S} produced is half of the moles of AgNO3\text{AgNO}_3: moles of Ag2S=12×moles of AgNO3=12×0.0315 mol=0.01575 mol\text{moles of Ag}_2\text{S} = \frac{1}{2} \times \text{moles of AgNO}_3 = \frac{1}{2} \times 0.0315 \, \text{mol} = 0.01575 \, \text{mol}
- Convert moles of silver(I) sulfide to grams: The molar mass of silver(I) sulfide (Ag2S\text{Ag}_2\text{S}) is 247.80 g/mol. To find the mass of Ag2S\text{Ag}_2\text{S}, use the formula: mass of Ag2S=moles of Ag2S×molar mass of Ag2S\text{mass of Ag}_2\text{S} = \text{moles of Ag}_2\text{S} \times \text{molar mass of Ag}_2\text{S} mass of Ag2S=0.01575 mol×247.80 g/mol=3.90 g\text{mass of Ag}_2\text{S} = 0.01575 \, \text{mol} \times 247.80 \, \text{g/mol} = 3.90 \, \text{g}
Thus, the mass of silver(I) sulfide (Ag2S\text{Ag}_2\text{S}) that will form is 3.90 grams.
Explanation:
In this reaction, sodium sulfide (Na2S\text{Na}_2\text{S}) is present in excess, meaning it will not limit the reaction. The amount of silver(I) sulfide formed is determined by the amount of silver(I) nitrate (AgNO3\text{AgNO}_3) available. First, we calculate the number of moles of silver(I) nitrate based on its given mass. Using the stoichiometry from the balanced equation, we then determine how many moles of silver(I) sulfide will be produced, which is half the number of moles of AgNO3\text{AgNO}_3. Finally, by multiplying the moles of silver(I) sulfide by its molar mass, we calculate the mass of silver(I) sulfide formed. This method ensures we use all available silver(I) nitrate and excess sodium sulfide, yielding a precise mass of silver(I) sulfide.