Question

1. Tính thành phần phần trăm khối lượng của các nguyên tố trong hợp chất MgSO4; Al(NO3)3

giải giúp em đi ạ :((

Answer 1

\(\left\{{}\begin{matrix}\%Mg=\dfrac{24.1}{120}.100\%=20\%\\\%S=\dfrac{32.1}{120}.100\%=26,667\%\\\%O=\dfrac{16.4}{120}.100\%=53,333\%\end{matrix}\right.\)

\(\left\{{}\begin{matrix}\%Al=\dfrac{27.1}{213}.100\%=12,676\%\\\%N=\dfrac{14.3}{213}.100\%=19,718\%\\\%O=\dfrac{16.9}{213}.100\%=67,606\%\end{matrix}\right.\)

Answer 2

\(a.\%Mg=\dfrac{24\times100\%}{120}\)\(=20\%\)

\(\%S=\dfrac{32\times100\%}{120}\)\(=26,66\%\)\(\%\)

\(\%O=100\%-\left(20\%+26,66\%\right)\)\(=53,34\%\)

Answer 3

\(b.\%Al=\dfrac{27}{213}\)\(=12,67\%\)

\(\%N=\dfrac{42}{213}\)\(=19,71\%\)

\(\%O=100\%-12,67\%-19,71\%=67,62\%\)