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Câu 1:
\(m_{H_2S}=0,75.34=25,5(g)\\ m_{CaSO_4}=0,025.136=3,4(g)\\ m_{Fe_2O_3}=0,05.160=8(g)\)
Câu 2:
\(V_{N_2}=2,5.22,4=56(l)\\ V_{H_2}=0,03.22,4=0,672(l)\\ V_{O_2}=0,45.22,4=10,08(l)\\ V_{hh}=22,4.(0,2+0,25)=22,4.0,45=10,08(l)\)
a. \(m_{CO}=n_{CO}
.
M_{CO}=0,1
.
\left(12+16\right)=0,1
.
28=2,8\left(g\right)\)
\(V_{CO}=n_{CO}
.
24,79=0,1
.
24,79=2,479\left(l\right)\)
b. \(m_{H_2S}=n_{H_2S}
.
M_{H_2S}=5
.
\left(1
.
2+32\right)=5
.
34=170\left(g\right)\)
\(V_{H_2S}=n_{H_2S}
.
24,79=5
.
24,79=123,95\left(l\right)\)
\(m_{CO}=n\cdot M=0,1\cdot\left(12+16\right)=2,8\left(g\right)\\ V_{CO\left(dktc\right)}=n\cdot22,4=0,1\cdot22,4=2,24\left(l\right)\)
\(m_{H_2S}=n\cdot M=5\cdot\left(2+32\right)=170\left(g\right)\\ V_{H_2S\left(dktc\right)}=n\cdot22,4=5\cdot22,4=112\left(l\right)\)
Câu 1
\(m_{HNO_3}=0,3.63=18,9\left(g\right)\)
\(m_{CuSO_4}=1,5.160=240\left(g\right)\)
\(m_{AlCl_3}=2.133,5=267\left(g\right)\)
Câu 2
a) \(V_{N_2}=3.22,4=67,2\left(l\right)\)
\(V_{H_2}=0,45.22,4=10,08\left(l\right)\)
\(V_{O_2}=0,55.22,4=12,32\left(l\right)\)
b) \(V_{hh}=\left(0,25+0,75\right).22,4=22,4\left(l\right)\)
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
\(a_1,m_{CaCO_3}=0,25.100=25(g)\\ a_2,m_{SO_2}=\dfrac{3,36}{22,4}.64=9,6(g)\\ a_3,m_{H_2SO_4}=\dfrac{9.10^{23}}{6.10^{23}}.98=147(g)\)
V O 2 = n O 2 .22,4 = 0,05.22,4= 1,12(l)
V H 2 = n H 2 .22,4= 0,15.22,4= 3,36(l)
V C O 2 = n C O 2 .22,4=14.22,4 = 313,6(l)
\(V_{SO_2\left(dktc\right)}=n\cdot22,4=0,75\cdot22,4=16,8\left(l\right)\)
\(V_{H_2\left(dktc\right)}=n\cdot22,4=3\cdot22,4=67,2\left(l\right)\)
a. \(V_{SO_2}=n_{SO_2} . 24,79=0,75 . 24,79=18,5925\left(l\right)\)
b. \(V_{H_2}=n_{H_2} . 24,79=3 . 24,79=74,37\left(l\right)\)