Question

1)cho các CTHH sau. Tính % khối lượng mỗi nguyên tố đó:    a)CaCO3.           b)H2SO4.          C)Al2S3.                 d)CuO.           e)Fe2(SO4)3

Answer 1

a) Đặt n CaCO3 =1 ( mol )

 => \(\left\{{}\begin{matrix}n_{Ca}=1\left(mol\right)\\n_C=1\left(mol\right)\\n_O=3\left(mol\right)\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}m_{Ca}=40\left(g\right)\\m_C=12\left(g\right)\\m_O=48\left(g\right)\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\%m_{Ca}=40\%\\\%m_C=12\%\\\%m_O=48\%\end{matrix}\right.\)

b) Đặt n H2SO4 = 1 ( mol )

\(\Rightarrow\left\{{}\begin{matrix}n_H=2\left(mol\right)\\n_S=1\left(mol\right)\\n_O=4\left(mol\right)\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}m_H=2\left(g\right)\\m_S=32\left(g\right)\\m_O=64\left(g\right)\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\%m_H=2,04\%\\\%m_S=32,65\%\\\%m_O=65,31\%\end{matrix}\right.\)

c) Đặt n Al2S3 = 1 ( mol )

\(\Rightarrow\left\{{}\begin{matrix}n_{Al}=2\left(mol\right)\\n_S=3\left(mol\right)\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}m_{Al}=54\left(g\right)\\m_S=96\left(g\right)\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\%m_{Al}=36\%\\\%m_S=64\%\end{matrix}\right.\)

d) Đặt n CuO = 1 ( mol )

\(\Rightarrow\left\{{}\begin{matrix}n_{Cu}=1\left(mol\right)\\n_O=1\left(mol\right)\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}m_{Cu}=64\left(g\right)\\m_O=16\left(g\right)\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\%m_{Cu}=80\%\\\%m_O=20\%\end{matrix}\right.\)

e) Đặt n Fe2(SO4)3 = 1 ( mol )

\(\Rightarrow\left\{{}\begin{matrix}n_{Fe}=2\left(mol\right)\\n_S=3\left(mol\right)\\n_O=12\left(mol\right)\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}m_{Fe}=112\left(g\right)\\m_S=96\left(g\right)\\m_O=192\left(g\right)\end{matrix}\right.\Rightarrow}\left\{{}\begin{matrix}\%m_{Fe}=28\%\\\%m_S=24\%\\\%m_O=48\%\end{matrix}\right.\)

Answer 2

e) Đặt n Fe2(SO4)3 = 1 ( mol )

  \(\Rightarrow\left\{{}\begin{matrix}n_{Fe}=2\left(mol\right)\\n_S=3\left(mol\right)\\n_O=12\left(mol\right)\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}m_{Fe}=112\left(g\right)\\m_S=96\left(g\right)\\m_O=192\left(g\right)\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\%m_{Fe}=28\%\\\%m_S=24\%\\\%m_O=48\%\end{matrix}\right.\)