For the double-replacement reaction between silver nitrate and sodium bromide that produces silver bromide, we have:
a. If 4.50 mol of silver nitrate reacts, 463.02 grams of sodium bromide is required.
b. The mass of 844.97 g of silver bromide is formed.
Double-replacement reaction
The double-replacement reaction between silver nitrate and sodium bromide is the following:
AgNO₃ + NaBr ⇄ AgBr + NaNO₃ (1)
a. Mass of sodium bromide
Knowing the number of moles of AgNO₃ and the 1:1 stoichiometric ratio between AgNO₃ and NaBr (eq 1), we can calculate the mass of NaBr as follows:
[tex]m_{NaBr} = \frac{1\:mol\:NaBr}{1\:mol\:AgNO_{3}}*n_{AgNO_{3}}*M_{NaBr}[/tex]
Where:
- [tex]n_{AgNO_{3}}[/tex]: is the number of moles of AgNO₃ = 4.50 moles
- [tex]M_{NaBr}[/tex]: is the molar mass of NaBr = 102.894 g/mol
Hence, the mass of sodium bromide is:
[tex]m_{NaBr} = \frac{1\:mol\:NaBr}{1\:mol\:AgNO_{3}}*4.50\:moles\:AgNO_{3}*\frac{102.894\:g\:NaBr}{1\:mol\:NaBr} = 463.02 \:g[/tex]
b. Mass of silver bromide
We can find the mass of silver bromide formed knowing that 4.50 moles of silver nitrate react with NaBr and the stoichiometric ratio between AgNO₃ and AgBr is 1:1 (eq 1), so:
[tex] m_{AgBr} = \frac{1\:mol\:AgBr}{1\:mol\:AgNO_{3}}*4.50\:moles\:AgNO_{3}*M_{AgBr} [/tex]
Where:
- [tex]M_{AgBr}[/tex]: is the molar mass of silver bromide = 187.77 g/mol
Then, the mass of silver bromide formed is:
[tex] m_{AgBr} = \frac{1\:mol\:AgBr}{1\:mol\:AgNO_{3}}*4.50\:moles\:AgNO_{3}*\frac{187.77 \:g\:AgBr}{1\:mol\:AgBr} = 844.97 \:g [/tex]
Learn more about the double-replacement reaction here:
brainly.com/question/5723822
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