a) $C_2H_5OH + 3O_2 \xrightarrow{t^o} 2CO_2 + 3H_2O$

b) $n_{C_2H_5OH} = \dfrac{4,6}{46} = 0,1(mol)$

$n_{O_2} = 3n_{C_2H_5OH} = 0,3(mol)$
$V_{O_2} = 0,3.22,4 = 6,72(lít)$

c)

Theo PTHH : 

$n_{CO_2} = 2n_{C_2H_5OH} = 0,2(mol) \Rightarrow V_{CO_2} = 0,2.22,4 = 4,48(lít)$

$n_{H_2O} = 3n_{C_2H_5OH} = 0,3(mol) \Rightarrow m_{H_2O} = 0,3.18 = 5,4(gam)$

Theo gt ta có: $n_{C_4H_{10}}=0,3(mol)$

$C_4H_{10}+\frac{13}{2}O_2\rightarrow 4CO_2+5H_2O$

Ta có: $n_{O_2}=1,95(mol)\Rightarrow V_{O_2}=43,68(l)$

\(n_{C_4H_{10}} = \dfrac{6,72}{22,4} = 0,3(mol)\\ C_4H_{10} + \dfrac{13}{2}O_2 \xrightarrow{t^o} 4CO_2 + 5H_2O\\ n_{O_2} = \dfrac{13}{2}n_{butan} = 1,95(mol)\\ \Rightarrow V_{O_2} = 1,95.22,4 = 43,68(lít)\)

PTHH:  2\(C_4H_{10}\) +  13\(O_2\)   ---> 8\(CO_2\)  +  10\(H_2O\)   (Lập và cân bằng phương trình)

            0,3 mol    1,95 mol      1,2 mol     1,5 mol

+ Số mol của \(C_4H_{10}\)

\(n_{C_4H_{10}}\) = \(\dfrac{V}{22,4}\) = \(\dfrac{6,72}{22,4}\) = 0,3 (mol)

+ Thể tích \(O_2\) cần dùng:

\(V_{O_2}\) = n . 22,4 = 1,95 . 22,4 = 43,68 (lít)

______________________________________

Có gì không đúng nhắn mình nha bạn :))

a, \(CH_4+2O_2\underrightarrow{t^o}CO_2+2H_2O\)

b, \(n_{CH_4}=\dfrac{28}{22,4}=1,25\left(mol\right)\)

\(n_{CO_2}=n_{CH_4}=1,25\left(mol\right)\Rightarrow m_{CO_2}=1,25.44=55\left(g\right)\)

c, \(n_{O_2}=2n_{CH_4}=2,5\left(mol\right)\Rightarrow V_{O_2}=2,5.22,4=56\left(l\right)\)

Mg+Cl₂-to>MgCl₂

0,25---0,25----0,25 mol

n _Mg=6\24=0,25 mol

m _MgCl2=0,25.95=23,75g

V_Cl2=0,25.22,4=5,6l

a) PTHH : Mg + Cl₂ -> MgCl₂

\(n_{Mg}=\dfrac{m}{M}=\dfrac{6}{24}=0,25\left(mol\right)\)

b) \(m_{MgCl}=59,5.0,25=14,875\left(g\right)\)

c) \(V_{Cl}=0,25.22,4=5,6\left(l\right)\)

\(a,PTHH:Mg+Cl_2\underrightarrow{t^o}MgCl_2\)

\(b,n_{Mg}=\dfrac{m}{M}=\dfrac{6}{24}=0,25\left(mol\right)\\ Theo.PTHH:n_{Cl_2}=n_{MgCl_2}=n_{Mg}=0,25\left(mol\right)\\ m_{MgCl_2}=n.M=0,25.95=23,75\left(g\right)\\ c,V_{Cl_2\left(đktc\right)}=n.22,4=0,25.22,4=5,6\left(l\right)\)

$a)PTHH:2Al+6HCl\to 2AlCl_3+3H_2$

$b)n_{Al}=\dfrac{5,4}{27}=0,2(mol)$

Theo PT: $n_{H_2}=1,5.n_{Al}=0,3(mol)$

$\Rightarrow V_{H_2}=0,3.22,4=6,72(lít)$

$c)$Theo PT: $n_{HCl}=3n_{Al}=0,6(mol)$

$\Rightarrow m_{HCl}=0,6.36,5=21,9(g)$

$d)PTHH:2H_2+O_2\xrightarrow{t^o}2H_2O$

Theo PT: $n_{H_2O}=n_{H_2}=0,3(mol)$

$\Rightarrow m_{H_2O}=0,3.18=5,4(g)$

2Al+6HCl->2AlCl₃+3H₂

0,2--0,6----------------0,3

2H₂+O₂-to>2H₂O

0,3-------------0,3

=>n _Al=\(\dfrac{5,4}{27}\)=0,2 mol

=>V_H2=0,3.22,4=6,72l

=>m _HCl=0,6.26,5=21,9g

=>m _H2O=0,3.18=5,4g

\(a,PTHH:2Al+6HCl\rightarrow2AlCl_3+3H_2\)

\(b,n_{Al}=\dfrac{m}{M}=\dfrac{5,4}{27}=0,2\left(mol\right)\\ Theo.PTHH:n_{H_2}=\dfrac{3}{2}.n_{Al}=\dfrac{3}{2}.0,2=0,3\left(mol\right)\\ V_{H_2\left(đktc\right)}=n.22,4=0,3.22,4=6,72\left(l\right)\)

\(c,Theo.PTHH:n_{HCl}=3.n_{Al}=3.0,2=0,6\left(mol\right)\\ m_{HCl}=n.M=0,6.36,5=21,9\left(g\right)\)

\(d,PTHH:2H_2+O_2\underrightarrow{t^o}2H_2O\\ Theo.PTHH:n_{H_2O}=n_{H_2}=0,3\left(mol\right)\\ m_{H_2O}=n.M=0,3.18=5,4\left(g\right)\)

a) PTHH: \(CH_4+2O_2\underrightarrow{t^o}CO_2+2H_2O\)

b+c)

Vì trong chất khí, tỉ lệ số mol cũng chính là tỉ lệ về thể tích 

\(\Rightarrow\left\{{}\begin{matrix}V_{CO_2}=V_{CH_4}=11,2\left(l\right)\\V_{O_2}=2V_{CH_4}=22,4\left(l\right)\end{matrix}\right.\)

\(n_{O_2}=\dfrac{1456:1000}{22,4}=0,065\left(mol\right)\\ a,2C_4H_{10}+13O_2\rightarrow\left(t^o\right)8CO_2+10H_2O\\ b,n_{CO_2}=\dfrac{8}{13}.0,065=0,04\left(mol\right)\\ n_{H_2O}=\dfrac{10}{13}.0,065=0,05\left(mol\right)\\ b,m_{sp}=m_{CO_2}+m_{H_2O}=44.0,04+18.0,05=2,66\left(g\right)\\ c,n_{C_4H_{10}}=\dfrac{2}{13}.0,065=0,01\left(mol\right)\\ V_{gas}=\dfrac{100}{80}.0,01.22,4=0,28\left(l\right)\)