Question

Bài 4. Tính thành phần phần trăm về khối lượng các nguyên tố trong hợp chất sau:

a, AlCl₃ b, Cu(OH)₂ c, Fe₂(SO₄)₃

Answer 1

a,

\(M_{AlCl3}=133,5\left(\frac{g}{mol}\right)\)

\(\Rightarrow\left\{{}\begin{matrix}\%_{Al}=\frac{27}{133,5}.100\%=20,22\%\\\%_{Cl}=100\%-20,22\%=79,78\%\end{matrix}\right.\)

b,

\(M_{Cu\left(OH\right)2}=98\left(\frac{g}{mol}\right)\)

\(\Rightarrow\left\{{}\begin{matrix}\%_{Cu}=\frac{64}{98}.100\%=65,3\%\\\%_O=\frac{32}{98}.100\%=32,65\%\\\%_H=100\%-\left(65,3\%+32,65\%\right)=2,05\%\end{matrix}\right.\)

c,

\(M_{Fe2\left(SO4\right)3}=400\left(\frac{g}{mol}\right)\)

\(\Rightarrow\left\{{}\begin{matrix}\%_{Fe}=\frac{112}{400}.100\%=28\%\\\%_S=\frac{96}{400}.100\%=24\%\\\%_O=100\%-\left(28\%+24\%\right)=48\%\end{matrix}\right.\)