If 25 g of AlBr3 was produced whenCsBr was added to the reaction vessel, what quantity (grams) of CsBr was added?

The question is incomplete, the complete question is;

3CsBr + Al(OH)3 → AlBr3 + 3CsOH

If 25 g of AlBr3  was produced when 125 mL of a CsBr solution was added, what was the molarity of the CsBr solution?

2.25 M

Explanation:

Number of moles of AlBr3   produced = 25g/266.69 g/mol = 0.0937 moles of AlBr3  .

Now, from the reaction equation;

3 moles of CsBr yields 1 mole of AlBr3

Therefore, x  moles of CsBr  will yield 0.0937 moles of AlBr3  

x = 3 * 0.0937 = 0.2811 moles

If 125 mL of a CsBr solution was added

And;

n =CV

where;

n = number of moles

C = concentration of solution

V = volume of solution

C = n/V

C = 0.2811 moles * 1000/125

C= 2.25 M

neutrons are in the middle of the atom and the electrons, which are much smaller.

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