a) + n_Na = 11,5/23 = 0,5 (mol)

+ n_O2 = 1,2/24 = 0,05 (mol)

b) + m_Mg = 0,6.24 = 14,4 (g)

+ n_CO2 = \(\frac{1,8.10^{21}}{6.10^{23}}=0,003\left(mol\right)\)

c) V_CO2 = 0,175.22,4 = 3,92 (l)

V_H2 = 0,2.22,4 = 4,48 (l)

A) n_Na= \(\frac{m_{Na}}{M_{Na}}=\frac{11,5}{23}=0,5\left(mol\right)\)

\(n_{O_2}=\frac{V_{O_2}}{24}=\frac{1,2}{24}=0,05\left(mol\right)\)

TK

Từ C₂H₄O₂ ta có: M = 60 g/mol; m_C = 2 x 12 = 24 g; m_H = 4 x 1 = 4 g;

M_O = 2 x 16 = 32 g.

%C = (24 : 60) x 100% = 40%; %H = (4 : 60) x 100% = 6,67%;

%O = 100% - 40% - 6,67% = 53,33%.

a) C₂H₅OH + 3O₂ --t^o--> 2CO₂ + 3H₂O

 b) \(n_{C_2H_5OH}=\dfrac{2,3}{46}=0,05\left(mol\right)\)

PTHH: C₂H₅OH + 3O₂ --t^o--> 2CO₂ + 3H₂O

               0,05-->0,15-------->0,1

=> V_CO2 = 0,1.22,4 = 2,24 (l)

c) V_O2 = 0,15.22,4 = 3,36 (l)

=> V_kk = 3,36 : 20% = 16,8 (l)

\(n_{Al}=0,2mol\)

\(4Al+3O_2\underrightarrow{t^o}2Al_2O_3\)

0,2      0,15

\(V_{O_2}=0,15\cdot22,4=3,36l\)

Chọn D

16 gam khí oxi: n = m : M = 16 : 32 = 0,5 (mol)

V = n.22,4 = 0,5.22,4 = 11,2 (lít)

\(n_{CO_2}=\dfrac{5,6}{22,4}=0,25mol\)

\(\left\{{}\begin{matrix}CH_4:x\left(mol\right)\\C_2H_4:y\left(mol\right)\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}16x+28y=3,6\\BTC:x+2y=n_{CO_2}=0,25\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=0,05mol\\y=0,1mol\end{matrix}\right.\)

a)\(CH_4+2O_2\underrightarrow{t^o}CO_2+2H_2O\)

\(C_2H_4+3O_2\underrightarrow{t^o}2CO_2+2H_2O\)

b)\(m_{CH_4}=0,05\cdot16=0,8g\)

\(m_{C_2H_4}=0,1\cdot28=2,8g\)

c)\(\Sigma n_{O_2}=2n_{CH_4}+3n_{C_2H_4}=2\cdot0,05+3\cdot0,1=0,4mol\)

\(\Rightarrow V_{O_2}=0,4\cdot22,4=8,96l\)

\(\Rightarrow V_{kk}=5V_{O_2}=5\cdot8,96=44,8l\)

3. a) M_O2/M_N2 = 32/28 = 8/7

b) M_O2/M_CO = 32/28 = 8/7

c) M_O2/M_kk = 32/29

1 tính khối lượng của

a) 0.5 mol Fe2O3

\(M_{Fe_2O_3}=2\times56+3\times16=160\) (g/mol)

\(m_{Fe_2O_3}=n_{Fe_2O_3}\times M_{Fe_2O_3}=0,5\times112=56\left(g\right)\)

b) 0,15 mol CO2

\(M_{CO_2}=1\times12+2\times16=44\) (g/mol)

\(m_{CO_2}=n_{CO_2}\times M_{CO_2}=0,15\times44=6,6\left(g\right)\)

c) 5,6 lít O2 ( điều kiện tiêu chuẩn )

\(n_{O_2}=\frac{V_{O_2}}{22,4}=\frac{5,6}{22,4}=0,25\left(mol\right)\)

\(M_{O_2}=2\times16=32\) (g/mol)

\(m_{O_2}=n_{O_2}\times M_{O_2}=0,25\times32=8\left(g\right)\)

d) 8,96 lít H2 ( điều kiện tiêu chuẩn)

\(n_{H_2}=\frac{V_{H_2}}{22,4}=\frac{8,96}{22,4}=0,4\left(mol\right)\)

\(M_{H_2}=2\times1=2\) (g/mol)

\(m_{H_2}=n_{H_2}\times M_{H_2}=0,4\times2=0,8\left(g\right)\)

2 tính thể tích ( điều kiện tiêu chuẩn)

a) 0,125 mol Cl2

\(V_{Cl_2}=22,4\times n_{Cl_2}=22,4\times0,125=2,8\left(l\right)\)

b) 2,5 mol CH4

\(V_{CH_4}=22,4\times n_{CH_4}=22,4\times2,5=56\left(l\right)\)

c) 6,4 gam 02

\(M_{O_2}=2\times16=32\) (g/mol)

\(n_{O_2}=\frac{m_{O_2}}{M_{O_2}}=\frac{6,4}{32}=0,2\left(mol\right)\)

\(V_{O_2}=22,4\times n_{O_2}=22,4\times0,2=4,48\left(l\right)\)

d) 5,6 gam N2

\(M_{N_2}=2\times14=28\) (g/mol)

\(n_{N_2}=\frac{m_{N_2}}{M_{N_2}}=\frac{5,6}{28}=0,2\left(mol\right)\)

\(V_{N_2}=22,4\times n_{N_2}=22,4\times0,2=4,48\left(l\right)\)

3 tính tỉ khối của khí O2 so với

a) khí N2

\(d_{O_2;N_2}=\frac{M_{O_2}}{M_{N_2}}=\frac{2\times16}{2\times14}=\frac{8}{7}\)

b) khí CO

\(d_{O_2;CO}=\frac{M_{O_2}}{M_{CO}}=\frac{2\times16}{1\times12+1\times16}=\frac{8}{7}\)

c) không khí

\(d_{O_2;kk}=\frac{M_{O_2}}{M_{kk}}=\frac{2\times16}{29}=\frac{32}{29}\)

1. a) m_Fe2O3 = 0,5.160 = 80 (g)

b) m_CO2 = 0,15.44 = 6,6 (g)

c) n_O2 = 5,6/22,4 = 0,25 (mol) => m_O2 = 0,25.32 = 8 (g)

d) n_H2 = 8,96/22,4 = 0,4 (mol) => m_H2 = 0,4.2 = 0,8 (g)