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Nimfa-mama [501]
3 years ago
12

You need to make an aqueous solution of 0.215 M aluminum bromide for an experiment in lab, using a 300 mL volumetric flask. How

much solid aluminum bromide should you add
Chemistry
1 answer:
Karo-lina-s [1.5K]3 years ago
3 0

Answer:

We have to add 17.2 grams of aluminium bromide

Explanation:

Step 1: Data given

Molarity of the aluminum bromide solution = 0.215 M

Volume = 300 mL = 0.300 L

Molar mass aluminium bromide = 266.69 g/mol

Step 2: Calculate moles Aluminium bromide

moles AlBr3 = volume * molarity

Moles AlBr3 = 0.300 L * 0.215 M

Moles AlBR3 = 0.0645 moles

Step 3: Calculate mass aluminium bromide

Mass aluminium bromide = moles AlBr3 * molar mass AlBr3

Mass aluminium bromide = 0.0645 moles * 266.69 g/mol

Mass aluminium bromide = 17.2 grams

We have to add 17.2 grams of aluminium bromide

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Answer- 400 grams of AlCl3 is the maximum amount of AlCl3 produced during the experiment.

Given - Number of moles of Al(NO3)3 - 4 moles
Number of moles of NaCl - 9 moles
Find - Maximum amount of AlCl3 produced during the reaction.
Solution - The complete reaction is - Al(NO3)3 + 3NaCl --> 3NaNO3 + AlCl3
To find the maximum amount of AlCl3 produced during the reaction, we need to find the limiting reagent.
Mole ratio Al(NO3)3 - 4/1 - 4
Mole ratio NaCl - 9/3 - 3
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