19) \(n_{SO_2}=\dfrac{V_{\left(đkc\right)}}{24,79}=\dfrac{7,437}{24,79}=0,3\left(mol\right)\)

20) \(n_{NO_2}=\dfrac{V_{\left(đkc\right)}}{24,79}=\dfrac{9,916}{24,79}=0,4\left(mol\right)\)

22) \(V_{CO_2\left(đkc\right)}=n.24,79=0,3.24,79=7,437\left(l\right)\)

23) \(V_{O_2\left(đkc\right)}=0,2.24,79=4,958\left(l\right)\)

24) \(\left\{{}\begin{matrix}n_{SO_2}=\dfrac{m}{M}=\dfrac{16}{64}=0,25\left(mol\right)\\n_{CO_2}=\dfrac{m}{M}=\dfrac{4,4}{44}=0,1\left(mol\right)\end{matrix}\right.\)

`=>` \(n_{hh}=n_{SO_2}+n_{CO_2}=0,25+0,1=0,35\left(mol\right)\)

`=>` \(V_{hh\left(đkc\right)}=n.24,79=0,35.24,79=8,6765\left(l\right)\)

25) \(n_{CO_2}=\dfrac{V_{\left(đkc\right)}}{24,79}=\dfrac{4,958}{24,79}=0,2\left(mol\right)\)

`=>` \(m_{CO_2}=n.M=0,2.44=8,8\left(g\right)\)

26) \(n_{N_2}=\dfrac{m}{M}=\dfrac{28}{28}=1\left(mol\right)\)

`=>` \(V_{N_2\left(đkc\right)}=n.24,79=1.24,79=24,79\left(l\right)\)

27) \(n_{CO_2}=\dfrac{m}{M}=\dfrac{22}{44}=0,5\left(mol\right)\)

`=>` \(V_{CO_2\left(đkc\right)}=n.24,79=0,5.24,79=12,395\left(l\right)\)

28) \(n_{NO_2}=\dfrac{V_{\left(đkc\right)}}{24,79}=\dfrac{14,874}{24,79}=0,6\left(mol\right)\)

`=>` \(m_{NO_2}=n.M=0,6.46=27,6\left(g\right)\)

29) \(n_{MgO}=\dfrac{m}{M}=\dfrac{24}{40}=0,6\left(mol\right)\)

`=>` \(\text{Số phân tử MgO}=n.6.10^{23}=0,6.6.10^{23}=3,6.10^{23}\left(\text{phân tử}\right)\)

\(M_{đường}=\dfrac{34,2}{0,1}=342\left(g/mol\right)\)

1 mol đường có: \(\left\{{}\begin{matrix}n_C=\dfrac{342.42,1\%}{12}=12\left(mol\right)\\n_H=\dfrac{342.6,43\%}{1}=22\left(mol\right)\\n_O=\dfrac{342-12.12-22}{16}=11\left(mol\right)\end{matrix}\right.\)

=> CTHH của đường là C₁₂H₂₂O₁₁

a) Ta có: \(d_{X/H_2}=8,5;M_{H_2}=2\left(g/mol\right)\)

`=>` \(M_X=8,5.2=17\left(g/mol\right)\)

Trong 1 mol X có \(\left\{{}\begin{matrix}n_N=\dfrac{17.82,98\%}{14}=1\left(mol\right)\\n_H=\dfrac{17-14}{1}=3\left(mol\right)\end{matrix}\right.\)

`=>` X là NH₃ 

b) Đổi 112 ml = 0,112 l

`=>` \(n_{NH_3}=\dfrac{V_{\left(đktc\right)}}{22,4}=\dfrac{0,112}{22,4}=0,005\left(mol\right)\)

`=>` \(\left\{{}\begin{matrix}n_N=n_{NH_3}=0,005\left(mol\right)\\n_H=3.n_{NH_3}=3.0,005=0,015\left(mol\right)\end{matrix}\right.\)

`=>` \(\left\{{}\begin{matrix}\text{Số nguyên tử N}=n.6.10^{23}=0,005.6.10^{23}=3.10^{21}\left(\text{nguyên tử}\right)\\\text{Số nguyên tử H}=n.6.10^{23}=0,015.6.10^{23}=9.10^{21}\left(\text{nguyên tử}\right)\end{matrix}\right.\)

Số oxi hóa của C trong HCN là `-4`

Số oxi hóa của Cl trong ClO₂ là `+4` (Đáng ra là \(ClO_2^-\) thì Cl có số oxi hóa là `+2` chứ nhỉ?)

a) Gọi \(\left\{{}\begin{matrix}n_{Fe_2O_3}=a\left(mol\right)\\n_{CuO}=b\left(mol\right)\end{matrix}\right.\left(ĐK:a,b>0\right)\)

=> 160a + 80b = 4 (1); \(n_{HCl}=0,13.1=0,13\left(mol\right)\)
PTHH: Fe₂O₃ + 6HCl ---> 2FeCl₃ + 3H₂O

             a-------->6a-------->2a

           CuO + 2HCl ---> CuCl₂ + H₂O

            b------>2b-------->b

b) n_HCl = 6a + 2b = 0,13 (2)
Từ (1), (2) => a = 0,015; b = 0,02

=> \(\dfrac{n_{FeCl_3}}{n_{CuCl_2}}=\dfrac{2.0,015}{0,02}=\dfrac{3}{2}\)

Ta có: \(\left\{{}\begin{matrix}n_{Al}=\dfrac{m}{M}=\dfrac{2,7}{27}=0,1\left(mol\right)\\n_{O_2}=\dfrac{V_{\left(đktc\right)}}{22,4}=\dfrac{33,6}{22,4}=1,5\left(mol\right)\end{matrix}\right.\)

PTHH: \(2Mg+O_2\xrightarrow[]{t^o}2MgO\)

            \(4Al+3O_2\xrightarrow[]{t^o}2Al_2O_3\)

Theo PTHH: \(n_{O_2}=\dfrac{1}{2}.n_{Mg}+\dfrac{3}{4}.n_{Al}\)

=> \(n_{Mg}=2.\left(1,5-\dfrac{3}{4}.0,1\right)=2,85\left(mol\right)\)

=> \(\left\{{}\begin{matrix}m_{Mg}=n.M=2,85.24=68,4\left(g\right)\\m_{O_2}=n.M=1,5.32=48\left(g\right)\end{matrix}\right.\)

ĐLBTKL: \(m_{MgO}+m_{Al_2O_3}=m_{Al}+m_{Mg}+m_{O_2}\)

=> \(m_{sản.phẩm}=m_{MgO}+m_{Al_2O_3}=2,7+68,4+48=119,4\left(g\right)\)

Bài 6:

a) \(m_{CuSO_4}=n.M=0,6.160=96\left(g\right)\)

b) \(n_{Cl_2}=\dfrac{\text{Số phân tử}}{6.10^{^{23}}}=\dfrac{3.10^{^{23}}}{6.10^{^{23}}}=0,5\left(mol\right)\)

`=>` \(m_{Cl_2}=n.M=0,5.71=35,5\left(g\right)\)

c) \(n_{CH_4}=\dfrac{V_{\left(đktc\right)}}{22,4}=\dfrac{11,2}{22,4}=0,5\left(mol\right)\)

=>`` \(m_{CH_4}=n.M=0,5.16=8\left(g\right)\)

d) \(m_{C_{12}H_{22}O_{11}}=1,5.342=513\left(g\right)\)

Bài 7:

a) \(n_{CO_2}=\dfrac{m}{M}=\dfrac{11}{44}=0,25\left(mol\right)\)

`=>` \(V_{CO_2\left(đktc\right)}=n.22,4=0,25.22,4=5,6\left(l\right)\)

b) \(n_{H_2S}=\dfrac{\text{Số phân tử}}{6.10^{23}}=\dfrac{9.10^{23}}{6.10^{23}}=1,5\left(mol\right)\)

`=>` \(V_{H_2S\left(đktc\right)}=n.22,4=1,5.22,4=33,6\left(l\right)\)

c) \(V_{Cl_2\left(đktc\right)}=n.22,4=0,7.22,4=15,68\left(l\right)\)

d) N phân tử H₂ có 6.10²³ phân tử H₂

`=>` \(n_{H_2}=\dfrac{\text{Số phân tử}}{6.10^{23}}=\dfrac{6.10^{23}}{6.10^{23}}=1\left(mol\right)\)

`=>` \(V_{H_2\left(đktc\right)}=n.22,4=1.22,4=22,4\left(l\right)\)

\(6a)m_{CuSO_4}=n.M=0,6.160=96\left(g\right)\)

\(b)n_{Cl_2}=\dfrac{\text{Số phân tử }Cl_2}{N}=\dfrac{3.10^{23}}{6.10^{23}}=0,5\left(mol\right)\)

\(m_{Cl_2}=n.M=0,5.71=35,5\left(g\right)\)

\(c)n_{CH_4}=\dfrac{V}{22,4}=\dfrac{11,2}{22,4}=0,5\left(mol\right)\)

\(m_{CH_4}=n.M=0,5.16=8\left(g\right)\)

\(d)m_{C_6H_{12}O_6}=n.M=1,5.180=270\left(g\right)\)

\(7a)n_{CO_2}=\dfrac{m}{M}=\dfrac{11}{44}=0,25\left(mol\right)\)

\(V_{CO_2}=n.22,4=0,25.22,4=5,6\left(l\right)\)

\(b)n_{H_2S}=\dfrac{\text{Số phân tử }H_2S}{N}=\dfrac{9.10^{23}}{6.10^{23}}=1,5\left(mol\right)\)

\(V_{H_2S}=n.22,4=1,5.22,4=33,6\left(l\right)\)

\(c)V_{Cl_2}=n.22,4=0,7.22,4=15,68\left(l\right)\)

\(d)n_{H_2}=\dfrac{\text{Số phân tử }H_2}{N}=\dfrac{6.10^{23}}{6.10^{23}}=1\left(mol\right)\)

\(V_{H_2}=n.22,4=1.22,4=22,4\left(l\right)\)

Bài 4:

Ta có: \(n_{CO_2}=\dfrac{\text{Số phân tử}}{6.10^{23}}=\dfrac{3.10^{23}}{6.10^{23}}=0,5\left(mol\right)\)

`=>` \(\left\{{}\begin{matrix}V_{CO_2\left(đktc\right)}=n.22,4=0,5.22,4=11,2\left(l\right)\\m_{CO_2}=n.M=0,5.44=22\left(g\right)\end{matrix}\right.\)

Bài 5: 

a) \(n_S=\dfrac{m}{M}=\dfrac{16}{32}=0,5\left(mol\right)\)

b) \(n_{Fe}=\dfrac{m}{M}=\dfrac{28}{56}=0,5\left(mol\right)\)

c) \(n_{CaO}=\dfrac{m}{M}=\dfrac{14}{56}=0,25\left(mol\right)\)

d) \(n_{CO_2}=\dfrac{m}{M}=\dfrac{66}{44}=1,5\left(mol\right)\)

e) \(n_C=\dfrac{\text{Số nguyên tử}}{6.10^{23}}=\dfrac{3.10^{23}}{6.10^{23}}=0,5\left(mol\right)\)

f) \(n_{H_2O}=\dfrac{\text{Số phân tử}}{6.10^{23}}=\dfrac{9.10^{23}}{6.10^{23}}=1,5\left(mol\right)\)

Bài 1: \(n_{H_2O}=\dfrac{\text{Số phân tử}}{6.10^{23}}=\dfrac{9.10^{23}}{6.10^{23}}=1,5\left(mol\right)\)

Bài 2: \(m_{SO_2}=n.M=0,5.64=32\left(g\right)\)

Bài 3: \(V_{H_2S\left(đktc\right)}=n.22,4=0,25.22,4=5,6\left(l\right)\)

\(1)n_{H_2O}=\dfrac{\text{Số phân tử }H_2O}{N}=\dfrac{9.10^{23}}{6.10^{23}}=1,5\left(mol\right)\)

\(2)m_{SO_2}=n.M=0,5.64=32\left(g\right)\)

\(3)V_{H_2S}=n.22,4=0,25.22,4=5,6\left(l\right)\)