$K_2SO_4.Al_2(SO_4)_3.24H_2O$
Viết lại CTHH của phèn chua:
$KAl(SO_4)_2.12H_2O$
$\Rightarrow M=39+27+96.2+18.12=474$
$\%m_K=\dfrac{39.100}{474}=8,23\%$
$\%m_{Al}=\dfrac{27.100}{474}=5,7\%$
$\%m_S=\dfrac{32.2.100}{474}=13,5\%$
$\%m_H=\dfrac{12.2.100}{474}=5,06\%$
$\%m_O=67,51\%$