1.

\(a,n_{Al}=\dfrac{5,4}{27}=0,2(mol)\\ n_{CuO}=\dfrac{8}{80}=0,1(mol)\\ n_{Fe_2O_3}=\dfrac{16}{160}=0,1(mol)\\ b,V_{CO_2}=0,25.22,4=5,6(l)\\ V_{H_2}=0,175.22,4=3,92(l)\\ V_{N_2}=1,5.22,4=33,6(l)\)

2.

\(a,n_{Al}=0,5.2=1(mol);n_{O}=0,5.3=1,5(mol)\\ \Rightarrow m_{Al}=1.27=27(g);m_{O}=1,5.16=24(g)\\ b,n_{CO_2}=\dfrac{2,2}{44}=0,05(mol)\\ \Rightarrow m_C=0,05.12=0,6(g);m_O=0,05.2.16=1,6(g)\)

\(a.\)

\(m_{CuO}=2\cdot40=80\left(g\right)\)

\(b.\)

\(n_{CO_2}=\dfrac{8.96}{22,4}=0.4\left(mol\right)\)

\(c.\)

\(n_{CH_4}=\dfrac{1.2\cdot10^{23}}{6\cdot10^{23}}=0.2\left(mol\right)\)

\(V_{CH_4}=0.2\cdot22.4=4.48\left(l\right)\)

a.\(m_{CO_2}=\dfrac{4,48}{22,4}.44=8,8g\)

b.\(n_{CH_4}=\dfrac{2,8}{22,4}=0,125mol\)

c.\(V_{Cl_2}=0,25.22,4=5,6l\)

a) nCo2= V/22,4= 4,48/22,4= 0,2(mol)                                                                   mCo2= n.M= 0,2.44=8.8(g)                                                                       b) nCH4= V/22,4= 2,8/22,4=0,125(mol)                                                           c) VCl2=n.22,4= 0,25.22,4=5,6(l)

m H 2 = n H 2 . M H 2  = 2.2 = 4(g) → V H 2 = n H 2 .22,4 = 2.22,4 = 44,8(l)

m O 2 = n O 2 . M O 2  = 1,5.32 = 48(g)  → V O 2 = n O 2 .22,4 = 1,5.22,4 = 33,6(l)

m C O 2 = n C O 2 . M C O 2  = 1,15.44 = 50,6(g)  → V C O 2 = n C O 2 .22,4 = 1,15.22,4 = 25,76(l)

m C H 4 = n C H 4 . M C H 4 = 1,15.16 = 18,4(g)  → V C H 4 = n C H 4 .22,4 = 1,15.22,4 = 25,76(l)

a) nCO2=[(9.10²³)/(6.10²³)]=1,5(mol)

=> mCO2=1,5.44=66(g)

V(CO2,đktc)=1,5.22,4=33,6(l)

b) nH2=4/2=2(mol)

N(H2)=2.6.10²³=12.10²³(phân tử)

V(H2,đktc)=2.22,4=44,8(l)

c) N(CO2)=0,5.6.10²³=3.10²³(phân tử)

V(CO2,đktc)=0,5.22,4=11,2(l)

mCO2=0,5.44=22(g)

d) nN2=2,24/22,4=0,1(mol)

mN2=0,1.28=2,8(g)

N(N2)=0,1.10²³.6=6.10²² (phân tử)

e) nCu=[(3,01.10²³)/(6,02.10²³)]=0,5(mol)

mCu=0,5.64=32(g)

Mà sao tính thể tích ta :3

\(a,m_{H_2S}=0,4.34=13,6(g);V_{H_2S}=0,4.22,4=8,96(l)\\ m_{SO_2}=0,025.64=1,6(g);V_{SO_2}=0,025.22,4=0,56(l)\\ m_{NO}=0,22.30=6,6(g);V_{NO}=0,22.22,4=4,928(l)\)

\(b,m_{CO}=0,45.28=12,6(g);V_{H_2S}=0,45.22,4=10,08(l)\\ m_{NH_3}=0,45.17=7,65(g);V_{NH_3}=0,45.22,4=10,08(l)\\ m_{CH_4}=0,45.16=7,2(g);V_{CH_4}=0,45.22,4=10,08(l)\\ m_{CO_2}=0,45.44=19,8(g);V_{CO_2}=0,45.22,4=10,08(l)\)

\(c,m_{hh}=0,1.28+0,3.48+0,375.36,5=30,8875(g)\\ V_{hh}=22,4.(0,1+0,3+0,375_17,36(l)\\ d,n_{O_2}=\dfrac{6.10^{23}}{6.10^{23}}=1(mol)\\ \Rightarrow m_{O_2}=32(g);V_{O_2}=22,4(l)\\ n_{N_2O_5}=\dfrac{7,2.10^{23}}{6.10^{23}}=1,2(mol)\\ \Rightarrow m_{N_2O_5}=1,2.108=129,6(g);V_{N_2O_5}=26,88(l)\\ n_{CO}=\dfrac{4,5.10^{23}}{6.10^{23}}=0,75(mol)\\ \Rightarrow m_{CO}=0,75.28=21(g);V_{CO}=0,75.22,4=16,8(l)\)

a,

\(mH_2S=0,4.34=13,6\left(gam\right)\):,\(VH_2S\left(đktc\right)=22,4.0,4=8,96lít\)

 \(mSO_2=0,025.64=1,6\left(gam\right)\);\(VSO_2=22,4.0,025=0,56l\)

M C H 4  = 12 + 4 = 16 g/mol

m C H 4 = n C H 4 . M C H 4  = 0,25.16 = 4(g)

m O 2 = n O 2 . M O 2  = 0,25.32 = 8(g)

m H 2 = n H 2 . M H 2  = 0,25. 2 = 0,5(g)

M C O 2  = 12 + 16.2 = 44 g/mol

m C O 2 = n C O 2 . M C O 2 = 0,25.44 = 11(g)

Ở đktc 0,25 mol của các chất khí đều có thể tích bằng nhau:

V C H 4 = V O 2 = V H 2 = V C O 2 = 22 , 4 . 0 , 25 = 5 , 6   ( l )

\(a_1,m_{CaCO_3}=0,25.100=25(g)\\ a_2,m_{SO_2}=\dfrac{3,36}{22,4}.64=9,6(g)\\ a_3,m_{H_2SO_4}=\dfrac{9.10^{23}}{6.10^{23}}.98=147(g)\)