Thank you for visiting The following reaction is carried out with 45 00 g of tex Na 2S tex and excess tex AgNO 3 tex What is the maximum. This page is designed to guide you through key points and clear explanations related to the topic at hand. We aim to make your learning experience smooth, insightful, and informative. Dive in and discover the answers you're looking for!
Answer :
To solve the problem of finding the maximum mass of [tex]\(Ag_2S\)[/tex] produced from the reaction, follow these steps:
1. Write down the balanced chemical equation:
[tex]\[
2 \, \text{AgNO}_3 + \text{Na}_2\text{S} \rightarrow \text{Ag}_2\text{S} + 2 \, \text{NaNO}_3
\][/tex]
From the equation, you can see that 1 mole of [tex]\( \text{Na}_2\text{S} \)[/tex] reacts to produce 1 mole of [tex]\( \text{Ag}_2\text{S} \)[/tex].
2. Find the molar masses:
- Molar mass of [tex]\( \text{Na}_2\text{S} \)[/tex] is approximately [tex]\( 78.04 \, \text{g/mol} \)[/tex].
- Molar mass of [tex]\( \text{Ag}_2\text{S} \)[/tex] is approximately [tex]\( 247.80 \, \text{g/mol} \)[/tex].
3. Calculate moles of [tex]\( \text{Na}_2\text{S} \)[/tex]:
Using the given mass of [tex]\( \text{Na}_2\text{S} \)[/tex]:
[tex]\[
\text{moles of Na}_2\text{S} = \frac{\text{mass of Na}_2\text{S}}{\text{molar mass of Na}_2\text{S}} = \frac{45.00 \, \text{g}}{78.04 \, \text{g/mol}} \approx 0.5766 \, \text{moles}
\][/tex]
4. Use stoichiometry to find moles of [tex]\( \text{Ag}_2\text{S} \)[/tex]:
According to the reaction, 1 mole of [tex]\( \text{Na}_2\text{S} \)[/tex] produces 1 mole of [tex]\( \text{Ag}_2\text{S} \)[/tex]. Therefore:
[tex]\[
\text{moles of Ag}_2\text{S} = 0.5766 \, \text{moles}
\][/tex]
5. Calculate the mass of [tex]\( \text{Ag}_2\text{S} \)[/tex]:
Use the moles of [tex]\( \text{Ag}_2\text{S} \)[/tex] to find its mass:
[tex]\[
\text{mass of Ag}_2\text{S} = \text{moles of Ag}_2\text{S} \times \text{molar mass of Ag}_2\text{S} = 0.5766 \, \text{moles} \times 247.80 \, \text{g/mol} \approx 142.89 \, \text{g}
\][/tex]
Therefore, the maximum mass of [tex]\( \text{Ag}_2\text{S} \)[/tex] produced is approximately 142.9 grams. The correct answer to the question is (A) 142.9 g.
1. Write down the balanced chemical equation:
[tex]\[
2 \, \text{AgNO}_3 + \text{Na}_2\text{S} \rightarrow \text{Ag}_2\text{S} + 2 \, \text{NaNO}_3
\][/tex]
From the equation, you can see that 1 mole of [tex]\( \text{Na}_2\text{S} \)[/tex] reacts to produce 1 mole of [tex]\( \text{Ag}_2\text{S} \)[/tex].
2. Find the molar masses:
- Molar mass of [tex]\( \text{Na}_2\text{S} \)[/tex] is approximately [tex]\( 78.04 \, \text{g/mol} \)[/tex].
- Molar mass of [tex]\( \text{Ag}_2\text{S} \)[/tex] is approximately [tex]\( 247.80 \, \text{g/mol} \)[/tex].
3. Calculate moles of [tex]\( \text{Na}_2\text{S} \)[/tex]:
Using the given mass of [tex]\( \text{Na}_2\text{S} \)[/tex]:
[tex]\[
\text{moles of Na}_2\text{S} = \frac{\text{mass of Na}_2\text{S}}{\text{molar mass of Na}_2\text{S}} = \frac{45.00 \, \text{g}}{78.04 \, \text{g/mol}} \approx 0.5766 \, \text{moles}
\][/tex]
4. Use stoichiometry to find moles of [tex]\( \text{Ag}_2\text{S} \)[/tex]:
According to the reaction, 1 mole of [tex]\( \text{Na}_2\text{S} \)[/tex] produces 1 mole of [tex]\( \text{Ag}_2\text{S} \)[/tex]. Therefore:
[tex]\[
\text{moles of Ag}_2\text{S} = 0.5766 \, \text{moles}
\][/tex]
5. Calculate the mass of [tex]\( \text{Ag}_2\text{S} \)[/tex]:
Use the moles of [tex]\( \text{Ag}_2\text{S} \)[/tex] to find its mass:
[tex]\[
\text{mass of Ag}_2\text{S} = \text{moles of Ag}_2\text{S} \times \text{molar mass of Ag}_2\text{S} = 0.5766 \, \text{moles} \times 247.80 \, \text{g/mol} \approx 142.89 \, \text{g}
\][/tex]
Therefore, the maximum mass of [tex]\( \text{Ag}_2\text{S} \)[/tex] produced is approximately 142.9 grams. The correct answer to the question is (A) 142.9 g.
Thank you for reading the article The following reaction is carried out with 45 00 g of tex Na 2S tex and excess tex AgNO 3 tex What is the maximum. We hope the information provided is useful and helps you understand this topic better. Feel free to explore more helpful content on our website!
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