Question

Tính thành phần phần trăm theo khối lượng các nguyên tố hoá học có trong các hợp chât sau: CO2,K3PO4,FeSO4,SO3,Al2O3,CuSO4,5H2O

Answer 1

CO₂ \(\left\{{}\begin{matrix}\%C=\dfrac{12}{44}.100\%=27,27\%\\\%O=100\%-27,27\%=72,73\%\end{matrix}\right.\)

K₃PO₄ \(\left\{{}\begin{matrix}\%K=\dfrac{117}{212}.100\%=55,19\%\\\%P=\dfrac{31}{212}.100\%=14,62\%\\\%O=100\%-55,19\%-14,62\%=30,19\%\end{matrix}\right.\)

FeSO₄ \(\left\{{}\begin{matrix}\%Fe=\dfrac{56}{152}.100\%=36,84\%\\\%S=\dfrac{32}{152}.100\%=21,05\%\\\%O=100\%-36,84\%-21,05\%=42,11\%\end{matrix}\right.\)

SO₃ \(\left\{{}\begin{matrix}\%S=\dfrac{32}{80}.100\%=40\%\\\%O=100\%-40\%=60\%\end{matrix}\right.\)

Al₂O₃ \(\left\{{}\begin{matrix}\%Al=\dfrac{54}{102}.100\%=52,94\%\\\%O=100\%-52,94\%=47,06\%\end{matrix}\right.\)

CuSO₄.5H₂O \(\left\{{}\begin{matrix}\%Cu=\dfrac{64}{250}.100\%=25,6\%\\\%S=\dfrac{32}{250}.100\%=12,8\%\\\%O=\dfrac{144}{250}.100\%=57,6\%\\\%H=100\%-25,6\%-12,8\%-57,6\%=4\%\end{matrix}\right.\)