\(n_{Ca\left(OH\right)2}=\frac{3,7}{74}=0,05\left(mol\right)\)
\(n_{Fe\left(OH\right)2}=\frac{49}{90}=0,5\left(mol\right)\)
\(n_{Al2\left(SO4\right)3}=\frac{171}{342}=0,5\left(mol\right)\)
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