A chemist diazotises 9.30 g of aniline (93 g/mol) in an ice bath with sodium nitrite (69 g/mol) and…
Contributed by StemPy LabTry it on paper first. A hint, the key idea and the worked solution are here when you need them.
A chemist diazotises of aniline () in an ice bath with sodium nitrite () and hydrochloric acid. The equation needs of hydrochloric acid for each mole of aniline. The acid is . Find the mass of sodium nitrite needed, the least amount of hydrochloric acid, and the volume of the acid that holds it. Then find the mass of benzenediazonium chloride () at a yield of 100 %.
Need a reminder? The key ideaShow
Diazotisation: , at 273 to 278 K. Each mole of salt that reacts to replace the diazonium group gives one mole of nitrogen gas, and one mole of gas is at 273 K and 1 atm. The yields of consecutive steps multiply.
The worked solution
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Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.
Step 1. Amount. .
Step 2. Sodium nitrite. .
Step 3. Acid. at least. In practice more is used, to keep the aniline as its salt.
Step 4. Volume. .
Step 5. The salt. One mole of salt for one mole of aniline: .
Answer: , , and .
Check. The salt holds the whole aniline unit plus and , so its mass must be above .
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