Hóa học

Bài 7:

Đặt CTHH là \(Ca_xN_yO_z\)

\(\%_O=100\%-24,4\%-17,1\%=58,5\%\\ x:y:z=\dfrac{24,4}{40}:\dfrac{17,1}{14}:\dfrac{58,5}{16}=0,61:1,22:3,66\approx1:2:6\\ \Rightarrow CTHH:Ca\left(NO_3\right)_2\)

Bài 8:

Đặt CTHH là \(C_xH_y\)

\(x:y=\dfrac{75}{12}:\dfrac{25}{1}=6,25:25=1:4\\ \Rightarrow CTHH:CH_4\)

\(n_B=\dfrac{44,8}{22,4}=2\left(mol\right)\)

=> \(M_B=\dfrac{4}{2}=2\left(g/mol\right)\)

=> B là H₂

\(d_{Cl_2/O_2}=\dfrac{35,5.2}{16.2}=2,21875\\ d_{Cl_2/kk}=\dfrac{35,5.2}{29}\approx2,45\)

a) \(n_{CO_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)

b) \(n_{CO_2}=\dfrac{8,8}{44}=0,2\left(mol\right)\)

c) \(n_{N_2}=\dfrac{5,6}{28}=0,2\left(mol\right)\)

Phân tử khối của H₂SO₄ là : 1.2+32+16.4=98 DvC

\(n_H=\dfrac{m}{M}=\dfrac{1.2}{98}=0,02mol\)

\(n_S=\dfrac{m}{M}=\dfrac{32}{98}=0,3mol\)

\(n_O=\dfrac{m}{M}=\dfrac{16.4}{98}=0,7mol\)

\(m_{H_2SO_4}=n.M=0,1.98=9,8\left(g\right)\)

A là Magie (M)

A là Magie (M)

\(M_A=\dfrac{m}{n}=\dfrac{4,8}{0,2}=24\left(\dfrac{g}{mol}\right)\)

Vậy A là nguyên tố magie (Mg)

 n = V/22,4.

Từ: \(V=n.22,4\Rightarrow n=\dfrac{V}{22,4}\)

Từ: \(m=n.M\Rightarrow n=\dfrac{m}{M}\)

a) \(\%Fe=\dfrac{56.3}{232}.100\%=72,41\%\)

b) \(\%Fe_2\left(SO_4\right)_3=\dfrac{56.2}{400}.100\%=28\%\)

Phân tử khối của Fe₃O₄ là : 56.3+16.4= 232 DvC

\(\%Fe=\dfrac{56.3}{232}.100\%=72\%\\ \%O=\dfrac{16.4}{232}.100\%=28\%\)

Phân tử khối của Fe₂(SO₄)₃=56.2+32.3+16.7=320 DvC

\(\%Fe=\dfrac{56.2}{320}.100\%=35\%\\ \%S=\dfrac{32.3}{320}.100\%=30\%\\ \%O=\dfrac{16.7}{320}.100\%=35\%\)

1) \(4K+O_2\rightarrow2K_2O\)

2) \(Mg+2HCl\rightarrow MgCl_2+H_2\uparrow\)

3) \(2Fe\left(OH\right)_3\underrightarrow{t^o}Fe_2O_3+3H_2O\)

4) \(2CO+O_2\rightarrow2CO_2\)