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?
Answer:
2.25 M
Step-by-step 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