Question

Tính thành phần % khối lượng các ng.tố trong hợp chất: Fe3O4 ; H2SO4.

(giúp mk với ;^;)

Answer 1

- \(\%_{Fe_{\left(Fe_3O_4\right)}}=\dfrac{56.3}{232}.100\%=72,4\%\)

\(\%_{O_{\left(Fe_3O_4\right)}}=100\%-72,4\%=27,6\%\)

- \(\%_{H_{\left(H_2SO_4\right)}}=\dfrac{1.2}{98}.100\%=2,04\%\)

\(\%_{S_{\left(H_2SO_4\right)}}=\dfrac{32}{98}.100\%=32,7\%\)

\(\%_{O_{\left(H_2SO_4\right)}}=100\%-2,04\%-32,7\%=65,26\%\%00\)

Answer 2

Fe₃O₄:

\(\%m_{Fe}=\dfrac{3.56}{232}.100\approx72,414\%\\ \Rightarrow\%m_O\approx27,586\%\\ \)

H₂SO₄:

\(\%m_H=\dfrac{2.1}{98}.100\approx2,041\%\\ \%m_S=\dfrac{32}{98}.100\approx32,653\%\\ \%m_O=\dfrac{4.16}{98}.100\approx65,306\%\)