Đáp án:+Giải thích các bước giải:

 Không khí= 29 đvC

a, Khí oxi: O2

$\frac{O2}{KK}$= $\frac{32}{29}$ ≈ 1,103

b, Khí cacbon oxit CO

$\frac{CO}{KK}$= $\frac{28}{29}$ ≈ 0,966

c, Khí sunfuro SO2

$\frac{SO2}{KK}$= $\frac{64}{29}$ ≈ 2,207

d, Khí cacbonic CO2

$\frac{CO2}{KK}$= $\frac{44}{29}$ ≈ 1,517

e, Khí hidro sunfua H2S

$\frac{H2S}{KK}$= $\frac{34}{29}$ ≈ 1,172

f, Khí nitơ: N2

$\frac{N2}{KK}$= $\frac{28}{29}$ ≈ 0,966

g, Khí ozon: O3

$\frac{O3}{KK}$= $\frac{48}{29}$ ≈ 1,655

h, Khí metan: CH4

$\frac{CH4}{KK}$= $\frac{16}{29}$ ≈ 0,552

l, Khí axetilen: C2H2

$\frac{C2H2}{KK}$= $\frac{26}{29}$ ≈ 0,897

j, Khí etilen C2H

$\frac{C2H}{KK}$= $\frac{25}{29}$ ≈ 0,962

a)

\({M_{{O_2}}} = 2{M_O} = 32 \to {d_{{O_2}/kk}} = \frac{{32}}{{29}}\)

b)

\({M_{CO}} = {M_C} + {M_O} = 12:16 = 28 \to {d_{CO/kk}} = \frac{{28}}{{29}}\)

c)

\({M_{C{O_2}}} = {M_C} + 2{M_O} = 12 + 16.2 = 44\)

\( \to {d_{C{O_2}/kk}} = \frac{{44}}{{29}}\)

d)

\({M_{S{O_2}}} = {M_S} + 2{M_O} = 32 + 16.2 = 64\)

\( \to {d_{S{O_2}/kk}} = \frac{{64}}{{29}}\)

e)

\({M_{{H_2}S}} = 2{M_H} + {M_S} = 2 + 32 = 34\)

\( \to {d_{{H_2}S/kk}} = \frac{{34}}{{29}}\)

f)

\({M_{{N_2}}} = 2{M_N} = 14.2 = 28 \to {d_{{N_2}/kk}} = \frac{{28}}{{29}}\)

g)

\({M_{{O_3}}} = 3{M_O} = 16.3 = 48\)

\( \to {d_{{O_3}/kk}} = \frac{{48}}{{29}}\)

h)

\({M_{C{H_4}}} = {M_C} + 4{M_{H}} = 12 + 4 = 16\)

\( \to {d_{C{H_4}/kk}} = \frac{{16}}{{29}}\)

i)

\({M_{{C_2}{H_2}}} = 2{M_C} + 2{M_H} = 12.2 + 2 = 26 \to {d_{{C_2}{H_2}/kk}} = \frac{{26}}{{29}}\)

j)

\({M_{{C_2}{H_4}}} = 2{M_C} + 4{M_H} = 12.2 + 4 = 28 \to {d_{{C_2}{H_4}/kk}} = \frac{{28}}{{29}}\)