a) Số mol của Fe là:

\(n_{Fe}=\dfrac{m_{Fe}}{M_{Fe}}=\dfrac{14}{56}=0,25\left(mol\right)\)

b) Thế tích khí của \(N_2\) là:

\(V_{N_2}=n_{N_2}\times22,4=0,15\times22,4=22,55\)

a, nFe=\(\dfrac{14}{56}\) = 0,25

b, V N2 (đktc) = 0,15 * 22.4 = 3,36 (l)

ta có: n_Cl2=\(\frac{7,1}{71}=0,1mol\)

\(V_{Cl2}=0,1.22,4=2,24\left(l\right)\)

\(n_{CO2}=\frac{8,8}{44}=0,2\left(mol\right)\)

\(V_{CO2}=0,2.22,4=4,48\left(l\right)\)

\(n_{NO2}=\frac{4,6}{46}=0,1\left(mol\right)\)

\(V_{NO2}=0,1.22,4=2,24\left(l\right)\)

\(n_{h^2}=0,1+0,2+0,1=0,4\left(mol\right)\)

\(V_{h^2}=2,24+2,24+4,48=8,96\left(l\right)\)

b) ta có \(n_{O2}=\frac{16}{32}=0,5\left(mol\right)\)

\(n_{N2}=\frac{14}{28}=0,5\left(mol\right)\)

\(\Leftrightarrow n_{h^2}=0,5+0,5=1\left(mol\right)\)

c) vì \(S=n.6.10^{23}\Rightarrow n=\frac{S}{6.10^{23}}\)

\(n_{N2}=\frac{1,5.10^{23}}{6.10^{23}}=0,25\left(mol\right)\)

\(V_{N2}=0,25.22,4=5,6\left(l\right)\)

\(n_{CO2}=\frac{9.10^{23}}{6.10^{23}}=1,5\left(mol\right)\)

\(V_{CO2}=1,5.22,4=33,6\left(l\right)\)

chúc bạn học tốt like mình nha

\(N_{Al}=0,25.6.10^{23}=1,5.10^{23}\)

\(n_{CO_2}=\dfrac{22}{44}=0,5mol\) 

\(\Rightarrow N_{CO_2}=0,5.6.10^{23}=3.10^{23}\)

\(n_{H_2}=\dfrac{3,36}{22,4}=0,15mol\)

\(\Rightarrow N_{H_2}=0,15.6.10^{23}=9.10^{22}\)

b. 

\(n_{N_2}=\dfrac{14}{28}=0,5\)

\(n_{Ca\left(NO_3\right)_2}=\dfrac{16,4}{164}=0,1mol\)

\(n_{SO_2}=\dfrac{1,12}{22,4}=0,05mol\)

a/ n_CO2 = 2,2 / 44 = 0,05 (mol)

=> V_CO2(đktc) = 0,05 x 22,4 = 1,12 lít

b/ n_Cl2 = 4,48 / 22,4 = 0,2 (mol)

=> m_Cl2 = 0,2 x 71 = 14,2 gam

c/ n_H2 = 0,5 / 2 = 0,25 (mol)

n_N2 = 14 / 28 = 0,5 (mol)

=> V_{hỗn hợp} = ( 0,25 + 0,5 ) x 22,4 = 16,8 lít

\(m_{Al_2O_3}=n.M=0,5.102=51\left(g\right)\\ m_{Fe_2O_3}=n.M=0,2.160=32\left(g\right)\\ \Rightarrow m_{hh}=51+32=83\left(g\right)\)

\(n_{hh}=0,2.0,5=0,7\left(mol\right)\\ \Rightarrow V_{hh}=n.22,4=0,7.22,4=15,68\left(l\right)\)

mAl2O3=0,5.(27.2+16.3)=51g

mFe2O3=0,2.(56.2+16.3)=32g

Vco2=0,2.22,4=4,48g

Vn2=0,5.22,4=11,2g

a.

\(V_{H_2S}=0.75\cdot22.4=16.8\left(l\right)\)

\(V_{SO_2}=\dfrac{12.8}{64}\cdot22.4=4.48\left(l\right)\)

\(V_{O_2}=\dfrac{3.2}{32}\cdot22.4=2.24\left(l\right)\)

b.

\(n_{hh}=\dfrac{22}{44}+\dfrac{3.55}{71}+\dfrac{0.14}{28}=0.555\left(mol\right)\)

\(V_{hh}=0.555\cdot22.4=12.432\left(l\right)\)

a) \(n_{Fe}=\dfrac{14}{56}=0,25\left(mol\right)\)

\(n_{CaCO_3}=\dfrac{25}{100}=0,25\left(mol\right)\)

\(n_{NaOH}=\dfrac{4}{40}=0,1\left(mol\right)\)

\(n_{...}=\dfrac{1,5.10^{23}}{6.10^{23}}=0,25\left(mol\right)\)

b) 

\(m_{ZnSO_4}=0,25.161=40,25\left(g\right)\)

\(m_{AlCl_3}=0,2.133,5=26,7\left(g\right)\)

\(m_{Cu}=0,3.64=19,2\left(g\right)\)

\(m_{Fe_2\left(SO_4\right)_3}=0,35.400=140\left(g\right)\)

d) \(V_{CO_2}=0,2.22,4=4,48\left(l\right)\)

\(V_{Cl_2}=0,15.22,4=3,36\left(l\right)\)

\(V_{SO_2}=0,3.22,4=6,72\left(l\right)\)

nX = 0,672/22,4 = 0,03 (mol)

Gọi nN2 = a (mol); nO2 = b (mol)

a + b = 0,03

28a + 32b = 0,88 

=> a = 0,02 (mol); b = 0,01 (mol)

%VN2 = 0,02/0,03 = 66,66%

%VO2 = 100% - 66,66% = 33,34%

M(X) = 0,88/0,03 = 88/3 (g/mol)

nX = 2,2 : 88/3 = 0,075 (mol)

VH2 = VX = 0,075 . 22,4 = 1,68 (l)

a)\(n_{Fe}=\dfrac{14}{56}=0,25\left(mol\right)\)

PTHH: Fe + 2HCl → FeCl₂ + H₂

Mol:    0,25    0,5                    0,25

b) \(C_{M_{ddHCl}}=\dfrac{0,5}{0,5}=1M\)

c) \(V_{H_2}=0,25.22,4=5,6\left(l\right)\)

a) 

\(n_{Fe}=\dfrac{14}{56}=0,25\left(mol\right)\)

\(n_{Ca}=\dfrac{20}{40}=0,5\left(mol\right)\)

\(n_{CaCO_3}=\dfrac{25}{100}=0,25\left(mol\right)\)

\(n_{NaOH}=\dfrac{4}{40}=0,1\left(mol\right)\)

\(n_{H_2O}=\dfrac{1,5.10^{23}}{6.10^{23}}=0,25\left(mol\right)\)

b) 

\(m_{ZnSO_4}=0,25.161=40,25\left(g\right)\)

\(m_{AlCl_3}=0,2.133,5=26,7\left(g\right)\)

\(m_{Cu}=0,3.64=19,2\left(g\right)\)

\(m_{Fe_2\left(SO_4\right)_3}=0,35.400=140\left(g\right)\)

c)

\(V_{CO_2}=0,2.22,4=4,48\left(l\right)\)

\(V_{Cl_2}=0,15.22,4=3,36\left(l\right)\)

\(V_{SO_2}=0,3.22,4=6,72\left(l\right)\)

\(V_{CH_4}=0,5.22,4=11,2\left(l\right)\)

a: \(n_{Fe}=\dfrac{14}{56}=0.25\left(mol\right)\)

\(n_{Ca}=\dfrac{20}{40}=0.5\left(mol\right)\)