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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
\(m_{CO_2}=12+16\cdot2=44\)
Số mol của CO2
\(n=\frac{m}{M}=\frac{11}{44}=0,25\)(mol)
Số mol của H2
\(n=\frac{sophantu}{6.10^{23}}=\frac{9.10^{23}}{6.10^{23}}=1,5\left(mol\right)\)
\(\Rightarrow\) VH2(đktc) =n.22,4=1,5.22,4=33,6(lít)
- nCO2 = 11 / 44 = 0,25 (mol)
- nH2 = \(\frac{9.10^{23}}{6.10^{23}}=1,5\left(mol\right)\)
=> VH2(đktc) = 1,5 x 22,4 = 33,6 lít
1.
\(a.\)
\(V_{hh}=\left(0.1+0.2+0.02+0.03\right)\cdot24=8.4\left(l\right)\)
\(b.\)
\(V_{hh}=\left(0.04+0.015+0.06+0.08\right)\cdot24=4.68\left(l\right)\)
\(2.\)
\(a.\)
\(V_{H_2}=0.5\cdot22.4=11.2\left(l\right)\)
\(V_{O_2}=0.8\cdot22.4=17.92\left(l\right)\)
\(b.\)
\(V_{CO_2}=2\cdot22.4=44.8\left(l\right)\)
\(V_{CH_4}=3\cdot22.4=67.2\left(l\right)\)
\(c.\)
\(V_{N_2}=0.9\cdot22.4=20.16\left(l\right)\)
\(V_{H_2}=1.5\cdot22.4=33.6\left(l\right)\)
\(a,n_{CO_2}=\dfrac{m_{CO_2}}{M_{CO_2}}=\dfrac{11}{44}=0,25\left(mol\right)\\ b,n_{H_2}=\dfrac{9.10^{23}}{6.10^{23}}=1,5\left(mol\right)\\ V_{H_2\left(đktc\right)}=1,5.22,4=33,6\left(l\right)\)
nSO2=m:M=6,4:64=0,1(mol)
VSO2=n.22,4=0,1.22,4=2,24(l)
nCO2=m:M=4,4:44=0,1(mol)
VCO2=n.22,4=0,1.22,4=2,24(l)
nH2=S:6.1023=1,2.1023:6.1023=0,2(mol)
VH2=n.22,4=0,2.22,4=4,48(l)
\(a.n_{SO_2}=\dfrac{m}{M}=\dfrac{6,4}{64}=0,1\left(mol\right)\\ n_{CO_2}=\dfrac{m}{M}=\dfrac{4,4}{44}=0,1\left(mol\right)\\ \Rightarrow n_{hh}=n_{SO_2}+n_{CO_2}=0,1+0,1=0,2\left(mol\right)\\ \Rightarrow V_{hh}=n.22,4=0,2.22,4=4,48\left(l\right)\)
\(n_{H_2}=\dfrac{1,2.10^{23}}{6.10^{23}}=0,2\left(mol\right)\Rightarrow V_{H_2}=n.22,4=0,2.22,4=4,48\left(l\right)\)