Answer :
To determine how many grams of AgBr are required to produce 75.0 grams of NaBr, we follow a series of logical steps based on stoichiometry and the provided molar masses.
1. Calculate the moles of NaBr:
- Given mass of NaBr: [tex]\( 75.0 \, \text{grams} \)[/tex]
- Molar mass of NaBr: [tex]\( 102.89 \, \text{g/mol} \)[/tex]
- Moles of NaBr can be calculated using the formula:
[tex]\[ \text{moles of NaBr} = \frac{\text{mass of NaBr}}{\text{molar mass of NaBr}} \][/tex]
Substituting the values:
[tex]\[ \text{moles of NaBr} = \frac{75.0 \, \text{grams}}{102.89 \, \text{g/mol}} \approx 0.7289338128097969 \, \text{moles} \][/tex]
2. Determine the moles of AgBr required:
- From the balanced chemical equation:
[tex]\[ 2 \, \text{AgBr} + \text{Na}_2\text{S}_2\text{O}_3 \rightarrow \text{Ag}_2\text{S}_2\text{O}_3 + 2 \, \text{NaBr} \][/tex]
The stoichiometric coefficients indicate that 2 moles of AgBr produce 2 moles of NaBr. Hence,
the number of moles of AgBr needed is the same as the number of moles of NaBr calculated.
Therefore,
[tex]\[ \text{moles of AgBr} = 0.7289338128097969 \, \text{moles} \][/tex]
3. Calculate the mass of AgBr required:
- Molar mass of AgBr: [tex]\( 187.77 \, \text{g/mol} \)[/tex]
- The mass of AgBr can be calculated using the formula:
[tex]\[ \text{mass of AgBr} = \text{moles of AgBr} \times \text{molar mass of AgBr} \][/tex]
Substituting the values:
[tex]\[ \text{mass of AgBr} = 0.7289338128097969 \, \text{moles} \times 187.77 \, \text{g/mol} \approx 136.87190203129558 \, \text{grams} \][/tex]
In conclusion, to produce 75.0 grams of NaBr, you would need approximately 136.87 grams of AgBr.
1. Calculate the moles of NaBr:
- Given mass of NaBr: [tex]\( 75.0 \, \text{grams} \)[/tex]
- Molar mass of NaBr: [tex]\( 102.89 \, \text{g/mol} \)[/tex]
- Moles of NaBr can be calculated using the formula:
[tex]\[ \text{moles of NaBr} = \frac{\text{mass of NaBr}}{\text{molar mass of NaBr}} \][/tex]
Substituting the values:
[tex]\[ \text{moles of NaBr} = \frac{75.0 \, \text{grams}}{102.89 \, \text{g/mol}} \approx 0.7289338128097969 \, \text{moles} \][/tex]
2. Determine the moles of AgBr required:
- From the balanced chemical equation:
[tex]\[ 2 \, \text{AgBr} + \text{Na}_2\text{S}_2\text{O}_3 \rightarrow \text{Ag}_2\text{S}_2\text{O}_3 + 2 \, \text{NaBr} \][/tex]
The stoichiometric coefficients indicate that 2 moles of AgBr produce 2 moles of NaBr. Hence,
the number of moles of AgBr needed is the same as the number of moles of NaBr calculated.
Therefore,
[tex]\[ \text{moles of AgBr} = 0.7289338128097969 \, \text{moles} \][/tex]
3. Calculate the mass of AgBr required:
- Molar mass of AgBr: [tex]\( 187.77 \, \text{g/mol} \)[/tex]
- The mass of AgBr can be calculated using the formula:
[tex]\[ \text{mass of AgBr} = \text{moles of AgBr} \times \text{molar mass of AgBr} \][/tex]
Substituting the values:
[tex]\[ \text{mass of AgBr} = 0.7289338128097969 \, \text{moles} \times 187.77 \, \text{g/mol} \approx 136.87190203129558 \, \text{grams} \][/tex]
In conclusion, to produce 75.0 grams of NaBr, you would need approximately 136.87 grams of AgBr.
