b, \(4P+5O_2\underrightarrow{t^o}2P_2O_5\)

\(n_P=\dfrac{6,2}{31}=0,2mol\\n_{O_2}=\dfrac{0,2.5}{4}=0,25mol \)

\(S+O_2\underrightarrow{t^o}SO_2\)

\(n_S=\dfrac{3,2}{32}=0,1mol\\ n_{O_2}=0,1mol\)

\(C+O_2\underrightarrow{t^o}CO_2\)

\(n_C=\dfrac{2,4}{12}=0,2mol\\ n_{O_2}=0,2mol\\ n_{O_2}\left(tổng\right)=\)

\(0,25+0,1+0,2=0,55mol\\ m_{O_2}\left(trong.hh.B\right)=0,55.32=17,6g\)

a, \(m_{Fe}=0,25.56=14g\)

\(3Fe+2O_2\underrightarrow{t^o}Fe_3O_4\)

\(\Rightarrow n_{O_2}=\dfrac{0,25.2}{3}=0,16mol\\ m_{O_2}=0,16.32=5,12g\)

\(4Al+3O_2\underrightarrow{t^o}2Al_2O_3\)

\(n_{O_2}=\dfrac{0,25.3}{4}=0,1875mol\\ m_{O_2}=0,1875.32=6g\)

\(2Zn+O_2\underrightarrow{t^o}2ZnO\)

\(n_{O_2}=\dfrac{0,5.1}{2}=0,25mol\\ m_{O_2}=0,25.32=8g\)

\(\Rightarrow m_{O_2}\left(trong.hỗn.hợp.A\right)=\) \(5,12+6+8=19,12g\)

help mi

nH2 = 8 : 2 = 4 (mol) 
nCH4 = 2 : 16 = 0,125 (mol) 
pthh : 2H2 + O2-t--> 2H2O 
         4------->2 (mol) 
       CH4 + 2O2 ----> CO2 + 2H2O 
        0,125->0,25 (MOL) 
=> nO2= 2 + 0,25 = 2,25 (mol) 
   => mO2 = 2,25 . 32 = 72 (g) 

C  +  O2 --t--> CO2 
0,15->0,15 (mol) 
S  +   O2 -t--> SO2 
0,125->0,125 (mol) 
=> nO2 = 0,15 + 0,125 = 0,275 (mol) 
=> mO2 = 0,275 . 32 = 8,8 (G)

Gọi \(\left\{{}\begin{matrix}n_{Mg}=x\left(mol\right)\\x_{Ca}=y\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}24x+40y=17,6\\x=2y\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,4\\y=0,2\end{matrix}\right.\)

a)\(m_{Mg}=0,4\cdot24=9,6g\)

   \(m_{Ca}=0,2\cdot40=8g\)

b)\(2Mg+O_2\underrightarrow{t^o}2MgO\)

   \(2Ca+O_2\underrightarrow{t^o}2CaO\)

Từ hai pt: \(\Rightarrow\Sigma n_{O_2}=\dfrac{1}{2}n_{Mg}+\dfrac{1}{2}n_{Ca}=\dfrac{1}{2}\cdot0,4+\dfrac{1}{2}\cdot0,2=0,3mol\)

\(\Rightarrow m_{O_2}=0,3\cdot32=9,6g\)

\(V_{O_2}=0,3\cdot22,4=6,72l\)

\(\Rightarrow V_{kk}=5V_{O_2}=5\cdot6,72=33,6l\)

a) 

Có \(\left\{{}\begin{matrix}24.n_{Mg}+40.n_{Ca}=17,6\\n_{Mg}=2.n_{Ca}\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}n_{Ca}=0,2\left(mol\right)\\n_{Mg}=0,4\left(mol\right)\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}m_{Ca}=0,2.40=8\left(g\right)\\m_{Mg}=0,4.24=9,6\left(g\right)\end{matrix}\right.\)

b)

PTHH: 2Ca + O₂ --t^o--> 2CaO 

            0,2-->0,1

             2Mg + O₂ --t^o--> 2MgO

            0,4--->0,2

=> \(V_{O_2}=\left(0,1+0,2\right).22,4=6,72\left(l\right)\)

\(V_{kk}=6,72.5=33,6\left(l\right)\)

\(a,Đặt:n_{CH_4}=a\left(mol\right);n_{C_4H_{10}}=b\left(mol\right)\left(a,b>0\right)\\ PTHH:CH_4+2O_2\rightarrow\left(t^o\right)CO_2+2H_2O\\ 2C_4H_{10}+13O_2\rightarrow\left(t^o\right)8CO_2+10H_2O\\ \Rightarrow\left\{{}\begin{matrix}16a+58b=7,4\\22,4a+22,4.4b=22\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,1\\b=0,1\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}m_{CH_4}=0,1.16=1,6\left(g\right)\\m_{C_4H_{10}}=0,1.58=5,8\left(g\right)\end{matrix}\right.\\ b,n_{O_2}=2a+\dfrac{13}{2}b=2.0,1+6,5.0,1=0,85\left(mol\right)\\ \Rightarrow V_{O_2\left(đktc\right)}=0,85.22,4=19,04\left(l\right)\)

 \(a,2H_2+O_2\rightarrow2H_2O\)

\(2CO+O_2\rightarrow2CO_2\)

Gọi \(a\) là số mol \(H_2\),\(b\) là số mol \(CO\)

\(n_{O_2}=\frac{89,6}{22,4}=4\left(mol\right)\)

Ta có:\(\hept{\begin{cases}2a+28b=68\\0,5+a+0,5b=4\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}a=6\\b=2\end{cases}}\)

\(b,m_{H_2}=6.2=12g\)

\(m_{CO}=2.28=56\left(g\right)\)

\(c,\%V_{H_2}=\frac{6}{8}.100\%=75\%\)

\(\%V_{CO}=100-75=25\%\)

Gọi x, y là số mol của C và S, ta có:

\(C+O_2\rightarrow CO_2\)

x       x            x

\(S+O_2\rightarrow SO_2\)

y       y          y

\(\Rightarrow\left\{{}\begin{matrix}x+y=\dfrac{4,48}{22,4}=0,2\\12x+32y=4,4\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=0,1\\y=0,1\end{matrix}\right.mol\)

\(\Rightarrow m_{hh}=0,1.44+0,1.64=10,8g\)

$n_{CO_2} = \dfrac{8,8}{44} = 0,2(mol)$

\(2CO+O_2\xrightarrow[]{t^o}2CO_2\)

0,2       0,1       0,2                 (mol)

$n_{O_2} = \dfrac{9,6}{32} = 0,3(mol)$

\(2H_2+O_2\xrightarrow[]{t^o}2H_2O\)

0,4     0,2         0,2              (mol)

\(\%m_{CO}=\dfrac{0,2.44}{0,2.44+0,4.2}.100\%=91,67\%\\ \%m_{H_2}=100\%-91,67\%=8,33\%\)

\(\%n_{CO}=\dfrac{0,2}{0,2+0,4}.100\%=33,33\%\\ \%n_{H_2}=100\%-33,33\%=66,67\%\)