Question

tính kl Al2(SO4)3 biết mO-mAl =27,6 g .mn ghi cả cách giải ra nhá ạ thanksss

Answer 1

Gọi số mol của Al₂(SO₄)₃ là a (mol)

\(\Rightarrow\left\{{}\begin{matrix}n_{Al}=2a\left(mol\right)\\n_O=12a\left(mol\right)\end{matrix}\right.\) \(\Rightarrow16\cdot12a-27\cdot2a=27,6\) \(\Leftrightarrow a=0,2\left(mol\right)\)

\(\Rightarrow m_{Al_2\left(SO_4\right)_3}=0,2\cdot342=68,4\left(g\right)\)

Answer 2

 

\(M_{Al_2\left(SO_4\right)_3}=2.M_{Al}+3.\left(NTK_S+4.NTK_O\right)\\ =2.27+3.\left(32+4.16\right)=342\left(\dfrac{g}{mol}\right)\)

\(\dfrac{m_O}{m_{Al}}=\dfrac{4.3.16}{27.2}=\dfrac{32}{9}\)

Mặt khác: m_O - m_Al= 27,6  => m_O=27,6+m_Al

=> \(\dfrac{27,6+m_{Al}}{m_{Al}}=\dfrac{32}{9}\Leftrightarrow m_{Al}=10,8\left(g\right)\\\Rightarrow n_{Al}=\dfrac{10,8}{27}=0,4\left(mol\right)\\ \Rightarrow n_{Al_2\left(SO_4\right)_3}=\dfrac{0,4}{2}=0,2\left(mol\right)\\ \Rightarrow m_{Al_2\left(SO_4\right)_3}=n_{Al_2\left(SO_4\right)_3}.M_{Al_2\left(SO_4\right)_3}=0,2.342=68,4\left(g\right)\)