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a)
b) VCO2 = 22,4 .0,175 = 3,92l.
VH2 = 22,4 .1,25 = 28l.
VN2 = 22,4.3 = 67,2l.
c) Số mol của hỗn hợp khí bằng tổng số mol của từng khí.
nhh = nCO2 + nH2 + nN2 = 0,01 + 0,02 + 0,02 = 0,05 mol
Vhh khí = (0,01 + 0,02 + 0,02) . 22,4 = 1,12l.
a)mFe=28g
MFe=56g
nFe=?
nFe=\(\dfrac{28}{56}\) =0,5mol
tương tự
nCu=\(\dfrac{64}{64}\) =1mol
nAl=\(\dfrac{5,4}{27}\) =0,2mol
b)tóm tắt
nCO2=0,175mol
V1molCO2=22,4(l)
V=?
VCO2=0,175x22,4=3,92(l)
tương tự
VH2=1,25x22,4=28(l)
VN2=3x22,4=67,2(l)
c)tóm tắt
mCO2=0,44g
MCO2=44g
V1molCO2=22,4(l)
nCO2=?
VCO2=?
nCO2=\(\dfrac{0,44}{44}\) 0,01,mol
VCO2=0,01x22,4=0,224(l)
tương tự
nH2=\(\dfrac{1,25}{2}\) 0,625mol
VH2=0,625x22,4=149(l)
nN2=\(\dfrac{0,56}{14}\) 0,04mol
VN2=0,04x22,4=0,896(l)
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) \(n_{Mg}=\dfrac{2,4}{24}=0,1\left(mol\right)\)
\(n_{Zn}=\dfrac{32,5}{65}=0,5\left(mol\right)\)
\(n_{Al}=\dfrac{2,7}{27}=0,1\left(mol\right)\)
\(n_{Cu}=\dfrac{19,2}{64}=0,3\left(mol\right)\)
b)
\(m_{CO_2}=44.0,5=22\left(g\right)\)
\(m_{H_2}=1,5.2=3\left(g\right)\)
\(m_{N_2}=2.28=56\left(g\right)\)
\(m_{CuO}=3.80=240\left(g\right)\)
c) \(n_{Cl_2}=\dfrac{14,2}{71}=0,2\left(mol\right)\)
\(n_{H_2}=\dfrac{4,8}{2}=2,4\left(mol\right)\)
\(n_{O_2}=\dfrac{3,2}{32}=0,1\left(mol\right)\)
=> nhh = 0,2 + 2,4 + 0,1 = 2,7 (mol)
=> Vhh = 2,7.22,4 = 60,48(l)
\(a.n_{Mg}=\dfrac{2,4}{24}=0,1\left(mol\right)\\ n_{Zn}=\dfrac{32,5}{65}=0,5\left(mol\right)\\ n_{Al}=\dfrac{2,7}{27}=0,1\left(mol\right)\\ n_{Cu}=\dfrac{19,2}{64}=0,3\left(mol\right)\)
\(b.m_{CO_2}=0,5.44=22\left(g\right)\\ m_{H_2}=1,5.2=3\left(g\right)\\ m_{N_2}=2.28=56\left(g\right)\\ m_{CuO}=3.80=240\left(g\right)\)
\(c.n_{Cl_2}=\dfrac{14,2}{71}=0,2\left(mol\right)\\ n_{H_2}=\dfrac{4,8}{2}=2,4\left(mol\right)\\ n_{O_2}=\dfrac{3,2}{32}=0,1\left(mol\right)\\\Rightarrow n_{hh}=0,2+2,4+0,1=2,7\left(mol\right)\\ \Rightarrow V_{hh}=2,7.22,4=60,48\left(l\right)\)
\(a,n_{Mg}=\dfrac{2,4}{24}=0,1(mol); n_{Zn}=\dfrac{32,5}{65}=0,5(mol)\\ n_{Al}=\dfrac{2,7}{27}=0,1(mol); n_{Cu}=\dfrac{19,2}{64}=0,3(mol)\)
\(b,m_{CO_2}=0,5.44=22(g);m_{H_2}=1,5.2=3(g)\\ m_{N_2}=2.28=56(g);m_{CuO}=3.80=240(g)\)
\(c,n_{hh}=n_{Cl_2}+n_{H_2}+n_{O_2}=\dfrac{14,2}{71}+\dfrac{4,8}{2}+\dfrac{3,2}{32}=0,2+2,4+0,1=2,7(mol)\\ V_{hh}=2,7.22,4=60,48(l)\)
a)
n(Fe)=28/56=0.5 mol
n(Cu)=64/64=1 mol
n(Al)=5.4/27=0.2 mol
b)
V(CO2)=0.175*22.4=3.92 lít
V(H2)=1.25*22.4=28 lít
V(N2)=3*22.4=67.2 lít
c)
n(CO2)=0.44/44=0.01 mol
n(H2)=0.04/2=0.02 mol
n(N2)=0.56/28=0.02 mol
V(hh)= (0.01+0.02+0.02)*22.4=1.12 lít
n Fe = 28/56= 0,5 mol
n cu = 6,4 / 64= 0,1 mol
n Al = 5,4 / 27 =0,2 mol
V CO2 = 0,75 . 22,4 =16,8 l
V H2 = 1,25 . 22,4 = 28 l
V N2 = 3 . 22,4 =67,2 l
\(a.n_{CO_2}=\dfrac{4,4}{44}=0,1\left(mol\right);n_{O_2}=\dfrac{0,6.10^{23}}{6.10^{23}}=0,1\left(mol\right)\\ V_{hh}=\left(0,5+1,5+0,1+0,1\right).22,4=49,28\left(l\right)\\ b.m_{hh}=0,5.28+1,5.2+4,4+0,1.32=24,6\left(g\right)\)
a, VN\(_2\) ( đktc ) = 0,5 . 22,4 = 11,2 lít
VH\(_2\) = 1,5 . 22,4 = 33,6 lít
\(n_{CO_2}=\dfrac{4,4}{44}=0,1\) ( mol )
=> \(V_{CO_2}=0,1.22,4=2,24\) ( lít )
\(n_{O_2}=\dfrac{0,6.10^{23}}{6.10^{23}}=0,1\) ( mol )
=> V\(O_2\) = 0,1 .22,4 = 2,24 lít
=> Vhh = 11,2 + 33,6 + 2,24 + 2,24 = 49,28 lít
b, \(m_{N_2}=0,5.28=14\) ( g )
\(m_{H_2}=1,5.2=3\) ( g )
\(m_{CO_2}=0,1.44=4,4\) ( g )
\(m_{O_2}=0,1.32=3,2\) (g)
\(m_{hh}=14+3+4,4+3,2=24,6\) ( g )
Câu 1:
\(d_{N_2/kk}=\dfrac{28}{29}\approx 0,97\\ d_{SO_2/kk}=\dfrac{80}{29}\approx 2,76\\ d_{NO_2/kk}=\dfrac{46}{29}\approx 1,59\)
Câu 2:
\(n_{CO_2}=\dfrac{0,44}{44}=0,01(mol);n_{N_2}=\dfrac{6.10^{23}}{6.10^{23}}=1(mol)\\ \Rightarrow V_{hh}=22,4(0,01+0,1+1)=24,864(l)\)
Câu 1:
\(d_{N_2/kk}=\dfrac{28}{29}=0,966\)
\(d_{SO_2/kk}=\dfrac{64}{29}=2,207\)
\(d_{NO_2/kk}=\dfrac{46}{29}=1,586\)
Câu 2:
\(n_{CO_2}=\dfrac{0,44}{44}=0,01\left(mol\right)\)
\(n_{N_2}=\dfrac{6.10^{23}}{6.10^{23}}=1\left(mol\right)\)
=> Vhh = (0,01+0,1+1).22,4 = 24,864(l)
ta có: nCl2=\(\frac{7,1}{71}=0,1mol\)
\(V_{Cl2}=0,1.22,4=2,24\left(l\right)\)
\(n_{CO2}=\frac{8,8}{44}=0,2\left(mol\right)\)
\(V_{CO2}=0,2.22,4=4,48\left(l\right)\)
\(n_{NO2}=\frac{4,6}{46}=0,1\left(mol\right)\)
\(V_{NO2}=0,1.22,4=2,24\left(l\right)\)
\(n_{h^2}=0,1+0,2+0,1=0,4\left(mol\right)\)
\(V_{h^2}=2,24+2,24+4,48=8,96\left(l\right)\)
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;
nAl=\(\dfrac{5,4}{27}=0,2\left(mol\right)\)
b;
VH2=22,4.1,25=28(lít)
VN2=3.22,4=67,2(lít)
c;
nCO2=\(\dfrac{0,44}{44}=0,01\left(mol\right)\)
nH2=\(\dfrac{0,04}{2}=0,02\left(mol\right)\)
nN2=\(\dfrac{0,56}{28}=0,02\left(mol\right)\)
VCO2=22,4.0,01=0,224(mol)
VN2=VH2=22,4.0,02=0,448(lít)
a. nAl = \(\dfrac{5.4}{27}=0.2mol\)
b. V H2 = 1.25*22.4 = 28l.
V N2 = 3*22.4 = 67.2 l.
c. nCO2 = \(\dfrac{0.44}{44}=0.01mol\)
V CO2 = 0.01*22.4 = 0.224l.
nH2 = \(\dfrac{0.04}{2}=0.02mol\)
V H2 = 0.02*22.4 = 0.448l
nN2 = \(\dfrac{0.56}{28}=0.02mol\)
V H2 = 0.02*22.4 = 0.448l.
Tóm tắt
mFe = 28 g.
M Fe = 56 g/mol
nFe = ? mol
Giải như của bạn là được rồi nhé.