a. \(n_{H_2}=\dfrac{9.10^{23}}{6.10^{23}}=1,5\left(mol\right)\)

PTHH : Fe₂O₃ + 3H₂ -> 2Fe + 3H₂O

           0,5          1,5         1

\(m_{Fe_2O_3}=0,5.160=80\left(g\right)\)

b. \(m_{Fe}=1.56=56\left(g\right)\)

nH2 = 9.10^23/6.10^23 = 1,5 (mol)

PTHH: Fe2O3 + 3H2 -> (t°) 2Fe + 3H2O

Mol: 0,5 <--- 1,5 ---> 1

mFe2O3 = 0,5 . 160 = 80 (g)

mFe = 1 . 56 = 56 (g)

Fe₂O₃+3H₂-to>2Fe+3H₂O

0,5-------1,5---------1 mol

n _H2=\(\dfrac{9,10^{23}}{6.10^{23}}\)=1,5 mol

=>m _Fe3O4=0,5.232=116g

=>m _Fe=1.56=56g

\(a,n_{H_2O}=\dfrac{1,8}{18}=0,1(mol)\\ n_{H_2}=\dfrac{5,6}{22,4}=0,25(mol)\\ n_{Cu}=\dfrac{9.10^{23}}{6.10^{23}}=1,5(mol)\\ m_{O_2}=2.32=64(g)\\ V_{O_2}=2.22,4=44,8(l)\)

\(a,n_{CO_2}=\dfrac{m_{CO_2}}{M_{CO_2}}=\dfrac{11}{44}=0,25\left(mol\right)\\ b,n_{H_2}=\dfrac{9.10^{23}}{6.10^{23}}=1,5\left(mol\right)\\ V_{H_2\left(đktc\right)}=1,5.22,4=33,6\left(l\right)\)

a)

n Fe₂O₃=8/(56✖ 2+16❌ 3)=0.05mol

b)nH₂=6.72/22.4=0.3mol

c) nH₂=9.10²³/6.10²³=1.5mol

VH₂=1,5✖ 22.4=33.6l

d)nO₂=3,2/32=0,1mol

➡ nN₂=0,4mol

mN₂=0,4✖ 28=11,2g

e)nFe₂(SO₄)₃=8/400=0,02mol

f)nH₂=(1,2✖ 10^23)/6✖ 10^23=0,2mol

nN₂=2,8/28=0,1mol

VN₂=0,1✖ 22,4=2,24l

VO₂=1,5✖ 22,4=33,6l

VH₂=0,1 ✖ 22,4=2,24l

V_X=2,24+2,24+33,6=38.08l

mO₂=1,5❌ 32=48g

mN2=0,1✖ 28=2,8g

mH2=0,1✖ 2=0,2g

mX=2,8+0,2+48=51g

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

\(a,\%Na=\dfrac{23}{85}.100\%=27,06\%\\ \%N=\dfrac{14}{85}.100\%=16,47\%\\ \%O=100\%-27,06\%-16,47\%=56,47\%\\ b,\%Al=\dfrac{54}{234}.100\%=27,1\%\\ \%C=\dfrac{36}{234}.100\%=15,4\%\\ \%O=100\%-27,1\%-15,4\%=57,5\%\)

\(c,\%N=\dfrac{28}{79}.100\%=35,4\%\\ \%H=\dfrac{4}{79}.100\%=5,1\%\\ \%O=100\%-35,4\%-5,1\%=59,5\%\)

\(M_{NaNO_3}=23+14+16.3=85\left(\dfrac{g}{mol}\right)\\ \Rightarrow\%m_{Na}=\dfrac{23.100\%}{85}=27\%\\ \%m_N=\dfrac{14.100\%}{85}=16,47\%\\ \Rightarrow\%m_O=100\%-\left(16,47\%+27\%\right)=56,53\)