\(\%m_S=\dfrac{32}{80}.100\%=40\%\)

\(\%m_O=\dfrac{3.16}{80}.100\%=60\%\)

%S= \(\dfrac{32}{80}x100\%=40\%\)

=> %O= 100%-40%=60%

\(M_{K_2SO_4}=39.2+32+16.4=174(g/mol)\\ \%_K=\dfrac{39.2}{174}.100\%=44,83\%\\ \%_S=\dfrac{32}{174}.100\%=18,39\%\\ \%_O=100\%-44,83\%-18,39\%=36,78\%\)

\(\%K=\dfrac{39.2}{174}.100\%=44,8\%\)

\(\%S=\dfrac{32}{174}.100\%=18,4\%\)

\(\%O=100\%-44,8\%-18,4\%=36,8\%\)

Câu 2:

Trong 1 mol X: \(\left\{{}\begin{matrix}n_{Ag}=\dfrac{170.63,53\%}{108}=1\left(mol\right)\\n_N=\dfrac{170.8,23\%}{14}=1\left(mol\right)\\n_O=\dfrac{170\left(100\%-63,53\%-8,23\%\right)}{16}=3\left(mol\right)\end{matrix}\right.\)

Vậy CTHH của X là \(AgNO_3\)

Câu 1:

\(a,\%_{Fe}=\dfrac{56}{180}\cdot100\%=31,11\%\\ \%_N=\dfrac{14\cdot2}{180}\cdot10\%=15,56\%\\ \%_O=100\%-31,11\%-15,56\%=53,33\%\\ b,\%_{N\left(N_2O\right)}=\dfrac{14\cdot2}{44}\cdot100\%=63,63\%\\ \%_{O\left(N_2O\right)}=100\%-63,63\%=36,37\%\\ \%_{N\left(NO\right)}=\dfrac{14}{30}\cdot100\%=46,67\%\\ \%_{O\left(NO\right)}=100\%-46,67\%=53,33\%\\ \%_{O\left(NO_2\right)}=\dfrac{16\cdot2}{46}\cdot100\%=69,57\%\\ \%_{N\left(NO_2\right)}=100\%-69,57\%=30,43\%\)

a) Với Fe₃O₄ thì Fe là 72,4% và O là 27,6%;
    Với Fe₂O₃ thì Fe là 70% và O là 30%
b) Với SO₂ thì S là 50% và O là 50%
    Với SO₃ thì S là 40% và O là 60%
c) m_Cu= \(\dfrac{80.80}{100}\)=64(g) ; m_O=\(\dfrac{80.20}{100}\)=16(g)
    n_Cu=\(\dfrac{64}{64}\)=1(mol) ; n_O=\(\dfrac{16}{16}\)=1(mol)
    Vậy CTHH của oxit đồng màu đen là: CuO
d) d_A/H2=\(\dfrac{Ma}{2}\)=17 => M_A=2.17=34(đvC)
   H =\(\dfrac{5,88.34}{100}\)\(\approx\)2(đvC) ; S =\(\dfrac{94,12.34}{100}\)\(\approx\)32
=> CTHH của chất khí A là SH₂

\(b.M_{NANO_3}=23.1+14.1+16.3=85\\ \%N=\dfrac{14.1}{85}.100\%=16,47\%\)

\(c.M_{NO_2}=14.1+6.1.2=46\\ \%N=\dfrac{14.1}{46}.100\%=30,43\%\)

a) Fe₂O₃----> \(\dfrac{56.2}{56.2+16.3}.100\%=70\%\\ =>100\%-70\%=30\%\)

b) Fe₃O₄---->\(\dfrac{56.3}{56.3+16.4}.100\%=72,41\%\\ =>100\%-72,41\%=27,59\%\)

c)SO₃---->\(\dfrac{32}{32+16.3}.100\%=40\%\\ =>100\%-40\%=60\%\)

a, \(\%m_{Fe}=\dfrac{56}{56+16}.100\%\approx77,78\%\)

\(\%m_O\approx22,22\%\)

b, \(\%m_N=\dfrac{14}{14+3.1}.100\%\approx82,35\%\)

\(\%m_H\approx17,65\%\)

c, \(\%m_N=\dfrac{14.2}{14.2+16.5}.100\%\approx25,93\%\)

\(\%m_O\approx74,07\%\)

d, \(\%m_{Ca}=\dfrac{40}{40+12+16.3}.100\%=40\%\)

\(\%m_C=\dfrac{12}{40+12+16.3}.100\%=12\%\)

\(\%m_O=48\%\)

e, \(\%m_C=\dfrac{12.12}{12.12+22+16.11}.100\%\approx42,11\%\)

\(\%m_H=\dfrac{22}{12.12+22+16.11}.100\%\approx6,43\%\)

\(\%m_O\approx51,46\%\)

f, \(\%m_S=\dfrac{32}{32+16.3}.100\%=40\%\)

\(\%m_O=60\%\)

Bạn tham khảo nhé!

a) \(M_{FeO}=56+16=72\left(\dfrac{g}{mol}\right)\)

\(\%Fe=\dfrac{56}{72}.100\%=77,7\%\)

\(\%O=100\%-77,7\%=22,3\%\)

Còn lại bạn làm như bài này nè

a)\(M_{FeO}=56+16=72đvC\)

   \(\%Fe=\dfrac{56}{72}\cdot100\%=77,78\%\)

   \(\%O=100\%-77,78\%=22,22\%\)

b)\(M_{NH_3}=14+3=17đvC\)

   \(\%N=\dfrac{14}{17}\cdot100\%=82,35\%\)

   \(\%O=100\%-82,35\%=17,65\%\)

c)\(M_{N_2O_5}=14\cdot2+5\cdot16=108đvC\)

   \(\%N=\dfrac{14\cdot2}{108}\cdot100\%=25,92\%\)

   \(\%O=100-25,92\%=74,08\%\)

d)\(M_{CaCO_3}=40+12+3\cdot16=100đvC\)

   \(\%Ca=\dfrac{40}{100}\cdot100\%=40\%\)

   \(\%C=\dfrac{12}{100}\cdot100\%=12\%\)

   \(\%O=100\%-\left(40\%+12\%\right)=48\%\)

Hai câu sau bạn làm tương tự nhé!

a, \(\%m_{Al}=\dfrac{54.100\%}{102}=53\%\)

\(\%m_O=100\%-53\%=47\%\)

b, \(\%m_{Ca}=\dfrac{40.100\%}{232}=17,24\%\)

\(\%m_H=\dfrac{2.100\%}{232}=0,86\%\)

\(\%m_P=\dfrac{62.100\%}{232}=26,72\%\)

\(\Rightarrow\%m_O=100\%-17,24\%-0,86\%-26,72\%=55,18\%\)

- Hợp chất SO₃

%S = M_S : M_SO3 .100% = 32 : 80 .100% = 40%

%O = 100% - 40% = 60%

- Hợp chất Fe₂O₃

%Fe = 2M_Fe : M_Fe2O3 . 100% = 2.56 : 160. 100% = 70%

%O = 100% - 70% = 30%

- Hợp chất CO₂

%C = M_C : M_CO2 . 100% = 12 : 44 . 100% = 27,3 %

%O = 100% - 27,3% = 72,7%

\(\begin{array}{l} *SO_3:\\ \%S=\dfrac{32}{32+16\times 3}\times 100\%=40\%\\ \%O=\dfrac{16\times 3}{32+16\times 3}\times 100\%=60\%\\ *Fe_2O_3:\\ \%Fe=\dfrac{56\times 2}{56\times 2+16\times 3}\times 100\%=70\%\\ \%O=\dfrac{16\times 3}{56\times 2+16\times 3}\times 100\%=30\%\\ *CO_2:\\ \%C=\dfrac{12}{12+16\times 2}\times 100\%=27,27\%\\ \%O=\dfrac{16\times 2}{12+16\times 2}\times 100\%=72,73\%\end{array}\)