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a. \(n_{O_2}=\frac{m}{M}=\frac{25,6}{32}=0,8mol\)
\(\rightarrow V_{O_2\left(ĐKT\right)}=0,8.24=19,2l\)
\(n_{CO_2}=\frac{m}{M}=\frac{25,6}{44}=0,6mol\)
\(\rightarrow V_{O_2\left(ĐKT\right)}=0,6.24=14,4l\)
\(\frac{V_{O_2}}{V_{CO_2}}=\frac{19,2}{14,4}=1,33\) lần
b. \(V_{hh}=V_{CO_2}+V_{O_2}=19,2+14,4=33,6l\)
\(A_{hh}=A_{O_2}+A_{CO_2}=\left(6.10^{23}.0,8\right)+\left(6.10^{23}.0,6\right)=0,84.10^{23}\) nguyên tử
c. \(n_{tb}=\frac{n_{hh}}{2}=\frac{0,8+0,6}{2}=0,7mol\)
\(a,m_C=10.36\%=3,6\left(kg\right)=3600\left(g\right)\\ n_C=\dfrac{3600}{12}=300\left(mol\right)\\ m_S=10-3,6=6,4\left(kg\right)=6400\left(g\right)\\ n_S=\dfrac{6400}{32}=200\left(mol\right)\\ C+O_2\rightarrow\left(t^o\right)CO_2\\ S+O_2\rightarrow\left(t^o\right)SO_2\\ n_{O_2\left(tổng\right)}=n_C+n_S=300+200=500\left(mol\right)\\ V_{O_2\left(tổng\right)\left(đktc\right)}=500.22,4=11200\left(l\right)\\ V_{kk}=\dfrac{100}{20}V_{O_2\left(tổng\right)\left(đktc\right)}=5.11200=56000\left(l\right)\\ b,V_{hh\left(CO_2,SO_2\left(đktc\right)\right)}=22,4.\left(n_C+n_S\right)=22,4.\left(300+200\right)=11200\left(l\right)\)
Vì: %mCH4 = 80%
\(\Rightarrow m_{CH_4}=25.80\%=20\left(g\right)\Rightarrow n_{CH_4}=\dfrac{20}{16}=1,25\left(mol\right)\)
\(\Rightarrow m_{H_2}=5\left(g\right)\Rightarrow n_{H_2}=\dfrac{5}{2}=2,5\left(mol\right)\)
PT: \(2H_2+O_2\underrightarrow{t^o}2H_2O\)
\(CH_4+2O_2\underrightarrow{t^o}CO_2+2H_2O\)
Theo PT: \(\Sigma n_{O_2}=\dfrac{1}{2}n_{H_2}+2n_{CH_4}=3,75\left(mol\right)\)
\(\Rightarrow V_{O_2}=3,75.22,4=84\left(l\right)\)
Mà: %VO2 = 20%
\(\Rightarrow V_{kk}=\dfrac{84}{20\%}=420\left(l\right)\)
Bạn tham khảo nhé!
\(m_{Al}=27,8.19,2\%=5,4\left(g\right)\\ m_{Fe}=27,8-5,4=22,4\left(g\right)\\ \rightarrow\left\{{}\begin{matrix}n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\\n_{Fe}=\dfrac{22,4}{56}=0,4\left(mol\right)\end{matrix}\right.\)
PTHH:
4Al + 3O2 --to--> 2Al2O3
0,2-->0,15------->0,1
3Fe + 2O2 --to--> Fe3O4
0,4-->4/15--------->2/15
\(\rightarrow\left\{{}\begin{matrix}V_{kk}=\left(0,15+\dfrac{4}{15}\right).22,4.5=\dfrac{140}{3}\left(l\right)\\m_{Cran}=0,1.102+\dfrac{2}{15}.232=\dfrac{617}{15}\left(g\right)\end{matrix}\right.\)
mAl=27,8.19,42%=5,4g
⇒nAl=\(\dfrac{5,4}{27}\)=0,2mol
⇒nFe=\(\dfrac{27,8-5,4}{56}\)=0,4mol
4Al+3O2to→2Al2O34
3Fe+2O2to→Fe3O4
⇒nO2=\(\dfrac{3}{4}\)nAl+\(\dfrac{2}{3}\)nFe=\(\dfrac{5}{12}\)mol
⇒Vkk=\(\dfrac{5}{12}\).22,4.5=46,67l
b,
mrắn=27,8+mO2=27,8+\(\dfrac{5}{12}\)32=41,1g
\(\%H_2=100-20-30=50\%\)
\(GS:n_Z=1\left(mol\right)\)
\(\Rightarrow n_{H_2}=0.5\left(mol\right),n_{CO_2}=0.2\left(mol\right),n_{H_2S}=0.3\left(mol\right)\)
\(\overline{M}=\dfrac{0.5\cdot2+0.2\cdot44+0.3\cdot34}{1}=20\left(\dfrac{g}{mol}\right)\)
\(d_{\dfrac{Z}{kk}}=\dfrac{20}{29}=0.7\)
Z nhẹ hơn không khí 0.7 (lần)
\(b.\)\(n_{O_2}=\dfrac{9.6}{32}=0.3\left(mol\right)\)
\(\Rightarrow n_Z=0.3\left(mol\right)\)
\(m_Z=\overline{M}\cdot n=20\cdot0.3=6\left(g\right)\)