CaCO3 có PTK =100g/mol

=> %Ca=\(\frac{40}{100}.100=40\%\)

%C=12:100.100=12%

%O=3.16:100.100=48%

H2SO4 có PTK =98

=> %H=2.1:98.100=2%

%S=32:98.100=32,7%

=> %O=100-2-32,7=65,3%

Fe2O3 có PTK =160

=> %Fe=56.2:160.100=70%

=> %O=100-70=30%

Mg(OH)2 có PTK: 58g/mol

=> %Mg=24:56.100=42,9%

%H=1.2:56.100=3,6%

=> %O=100-42,9-3,6=56,5%

Đáp án: d

Giả thích:

Ta có: \(M_{X_2\left(SO_4\right)_3}=400\)

\(\Rightarrow2M_X+3\left(32+16.4\right)=400\)

\(\Rightarrow M_X=56\left(g/mol\right)\)

Vậy: X là Fe.

Bạn tham khảo nhé!

\(CaCO_3\\ \%m_{Ca}=\dfrac{40}{40+12+3.16}.100=40\%\\ \%m_C=\dfrac{12}{40+12+16.3}.100=12\%\\ \Rightarrow\%m_O=100\%-\left(40\%+12\%\right)=48\%\\ H_2SO_4\\ \%m_H=\dfrac{2.1}{2.1+32+4.16}.100\approx2,041\%\\ \%m_S=\dfrac{32}{2.1+32+4.16}.100\approx32,653\%\\ \%m_O=\dfrac{4.16}{2.1+32+4.16}.100\approx65,306\%\\ Fe_2O_3\\ \%m_{Fe}=\dfrac{56.2}{56.2+16.3}.100=70\%\\ \Rightarrow\%m_O=100\%-70\%=30\%\)

CaCO₃
\(\%M_{\dfrac{Ca}{CaCO_3}}=\dfrac{40}{100}.100\%=40\%\)
\(\%M_{\dfrac{C}{CaCO_3}}=\dfrac{12}{100}.100\%=12\%\)
\(\%M_{\dfrac{O}{CaCO_3}}=100\%-\left(40\%+12\%\right)=48\%\)
H₂SO₄
\(\%M_{\dfrac{H_2}{H_2SO_4}}=\dfrac{2}{98}.100\%=2,04\%\)
\(\%M_{\dfrac{S}{H_2SO_4}}=\dfrac{32}{98}.100\%=32,65\%\)
\(\%M_{\dfrac{O}{H_2SO_4}}=100\%-\left(2,04\%+32,65\%\right)=65,31\%\)
Fe₂O3
\(\%M_{\dfrac{Fe}{Fe_2O_3}}=\dfrac{112}{160}.100\%=70\%\)
\(\%M_{\dfrac{O}{Fe_2O_3}}=100\%-70\%=30\%\)
 

`a,` \(K.L.P.T_{Fe_2O_3}=56.2+16.3=160< amu>.\) 

\(\%Fe=\dfrac{56.2.100}{160}=70\%\)

\(\%O=100\%-70\%=30\%\)

`b,`\(K.L.P.T_{CaCO_3}=40+12+16.3=100< amu>.\)

\(\%Ca=\dfrac{40.100}{100}=40\%\)

\(\%C=\dfrac{12.100}{100}=12\%\) 

\(\%O=100\%-40\%-12\%=48\%\)

`c,` \(K.L.P.T_{HCl}=1+35,5=36,5< amu>.\)

\(\%H=\dfrac{1.100}{36,5}\approx2,74\%\)

\(\%Cl=100\%-2,74\%=97,26\%\)

a: \(\%Fe=\dfrac{56\cdot2}{56\cdot2+16\cdot3}=70\%\)

=>%O=30%

b: \(\%Ca=\dfrac{40}{40+12+16\cdot3}=40\%\)

\(\%C=\dfrac{12}{100}=12\%\)

%O=100%-12%-40%=48%

c: %H=1/36,5=2,74%

=>%Cl=97,26%

là sao?

Chọn B (Na2SO4)