\(a,PTHH:3Fe+2O_2\xrightarrow{t^o}Fe_3O_4\\ b,n_{Fe}=\dfrac{5,6}{56}=0,1(mol);n_{O_2}=\dfrac{3,2}{32}=0,1(mol)\)

Vì \(\dfrac{n_{Fe}}{3}<\dfrac{n_{O_2}}{2}\) nên \(O_2\) dư

\(n_{O_2(dư)}=0,1-0,1.\dfrac{2}{3}=0,033(mol)\\ \Rightarrow m_{O_2(dư)}=0,033.32=1,056(mol)\\ c,n_{Fe_3O_4}=\dfrac{1}{3}n_{Fe}=0,033(mol)\\ \Rightarrow m_{Fe_3O_4}=0,033.232=7,656(g)\)

sửa lại hộ mình chỗ \(m_{O_2(dư)}=1,056(g)\)

$a) 4P + 5O_2 \xrightarrow{t^o} 2P_2O_5$

$n_P = \dfrac{6,2}{31} = 0,2(mol) ; n_{O_2} = \dfrac{7,84}{22,4} = 0,35(mol)$
$n_P : 4 = 0,05 < n_{O_2} :5 = 0,07$ nên $O_2$ dư

$n_{O_2\ pư} = \dfrac{5}{4}n_P = 0,25(mol)$
$\Rightarrow m_{O_2\ dư} = (0,35 - 0,25).32 = 3,2(gam)$

c) $n_{P_2O_5} = \dfrac{1}{2}n_P = 0,1(mol)$
$m_{P_2O_5} = 0,1.142 = 14,2(gam)$

\(n_P=\dfrac{6,2}{31}=0,2\left(mol\right)\\ n_{O_2}=\dfrac{7,84}{22,4}=0,35\left(mol\right)\\a, 4P+5O_2\rightarrow\left(t^o\right)2P_2O_5\\ V\text{ì}:\dfrac{0,35}{5}>\dfrac{0,2}{4}\Rightarrow O_2d\text{ư}\\ n_{O_2\left(d\text{ư}\right)}=0,35-\dfrac{5}{4}.0,2=0,1\left(mol\right)\\b, m_{O_2\left(d\text{ư}\right)}=0,1.32=3,2\left(g\right)\\ c,n_{P_2O_5}=\dfrac{n_P}{2}=\dfrac{0,2}{2}=0,1\left(mol\right)\\ m_{r\text{ắn}}=m_{P_2O_5}=142.0,1=14,2\left(g\right)\)

nH2 = 0,4/2 = 0,2 (mol)

nO2 = 3,2/32 = 0,1 (mol)

PTHH: 2H2 + O2 -> (t°) 2H2O

LTL: 0,2/2 = 0,1 => khối lượng chất dư là 0

nH2O = nH2 = 0,2 (mol)

mH2O = 0,2 . 18 = 3,6 (g)

\(n_{H_2}=\dfrac{m_{H_2}}{M_{H_2}}=\dfrac{0,4}{2}=0,2mol\)

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

\(2H_2+O_2\rightarrow\left(t^o\right)2H_2O\)

 2         1                2      ( mol )

0,2   = 0,1                      ( mol )

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

Sau phản ứng ko có chất dư

\(m_{H_2O}=n_{H_2O}.M_{H_2O}=0,2.18=3,6g\)

a. \(n_{H_2}=\dfrac{0.4}{2}=0,2\left(mol\right)\)

\(n_{O_2}=\dfrac{3.2}{32}=0,1\left(mol\right)\)

PTHH : 2H₂ + O₂ ---t^o---> 2H₂O

              0,2     0,1              0,2

b. Ta thấy : 0,2 > 0,1 => H₂ dư , O₂ đủ

\(m_{H_2\left(dư\right)}=\left(0,2-0,2\right).2=0,2\left(g\right)\)

c. \(m_{H_2O}=0,2.18=3,6\left(g\right)\)

4Al+3O₂-to>2Al₂O₃

0,4----0,3------0,2

n _Al=\(\dfrac{10,8}{27}\)=0,4 mol

n _O2=\(\dfrac{7,84}{22,4}\)=0,35 mol

=> oxi dư

=>m _Al2O3=0,2.102=20,4g

=>m _{O2 dư}=0,05.32=1,6g

bạn ơi giúp mình với mình cần gấp

\(n_{Al}=\dfrac{10,8}{27}=0,4\left(mol\right)\)

\(n_{O_2}=\dfrac{7,84}{22,4}=0,35\left(mol\right)\)

PTHH: 4Al + 3O₂ --t^o--> 2Al₂O₃

Xét tỉ lệ: \(\dfrac{0,4}{4}< \dfrac{0,35}{3}\)=> Al hết, O₂ dư

PTHH: 4Al + 3O₂ --t^o--> 2Al₂O₃

            0,4-->0,3------->0,2

=> \(m_{Al_2O_3}=0,2.102=20,4\left(g\right)\)

\(V_{O_2\left(dư\right)}=\left(0,35-0,3\right).22,4=1,12\left(l\right)\)

a)\(n_{Mg}=\dfrac{35,6}{24}=1,483\left(mol\right)\)

  \(V_{O2\left(đktc\right)}=\dfrac{21,504}{22,4}=0,96\left(mol\right)\)

pt:    2Mg + O₂  →  2MgO (1)

mol:   2         1             2

mol:1,483    0,96

Tỉ lệ: \(\dfrac{1,483}{2}=0,7415< \dfrac{0,96}{1}=0,96\)

Mg tác dụng hết. O₂ dư

theo PTHH có

\(n_{O2p\intư}=\dfrac{1,843x1}{2}=0,7415\left(mol\right)\)

n_O2 dư=1,843-0,7415=1,1015 (mol)

m_O2dư= 1,1015 x 32 = 35,48 (g)

b)theo PTHH có

 \(n_{MgO}=\dfrac{1,843x2}{2}=1,843\left(mol\right)\)

n_MgO = 1,843 X 40 = 73,72 (g)

c)

n_{Mg PT(1)}=n_MgPT(2)=1,843 (mol)

pt:    Mg + H₂SO₄ ➝ MgSO₄ + H₂   (2)

mol:  1           1              1           1

mol:   1,843

Theo PTHH có

\(n_{H2}=\dfrac{1,843x1}{1}=1,843\) (mol)

m_H2=1,843 x 2 = 3,686 (g)

Đặt \(\left\{{}\begin{matrix}n_{Fe}=x\left(mol\right)\\n_{Fe_2O_3}=y\left(mol\right)\end{matrix}\right.\Rightarrow56x+160y=4,8\left(1\right)\)

\(PTHH:Fe+CuSO_4\rightarrow FeSO_4+Cu\\ \Rightarrow n_{Cu}=n_{Fe}=a\left(mol\right)\\ Fe_2O_3+6HCl\rightarrow2FeCl_3+3H_2O\\ Cu+2FeCl_3\rightarrow CuCl_2+2FeCl_2\\ \Rightarrow n_{Cu}=\dfrac{1}{2}n_{FeCl_3}=n_{Fe_2O_3}=b\left(mol\right)\\ \Rightarrow n_{Cu\left(dư\right)}=a-b=\dfrac{3,2}{64}=0,05\left(mol\right)\left(2\right)\)

\(\left(1\right)\left(2\right)\Rightarrow\left\{{}\begin{matrix}a=\dfrac{8}{135}\\b=\dfrac{1}{108}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\%_{Fe}=\left(\dfrac{8}{135}\cdot56\right):4,8\cdot100\%\approx69,14\%\\\%_{Fe_2O_3}\approx30,86\%\end{matrix}\right.\)

\(b,n_{HCl}=6n_{Fe_2O_3}=\dfrac{1}{18}\approx0,06\left(mol\right)\\ \Rightarrow V_{dd_{HCl}}=\dfrac{0,06}{1}=0,06\left(l\right)\)

\(a,n_{Al}=\dfrac{10,8}{27}=0,4\left(mol\right)\\ n_{O_2}=\dfrac{13,44}{22,4}=0,6\left(mol\right)\)

PTHH: 4Al + 3O₂ --t^o--> 2Al₂O₃

b, LTL: \(\dfrac{0,4}{4}>\dfrac{0,6}{3}\) => O₂ dư

Theo pthh: \(\left\{{}\begin{matrix}n_{O_2\left(pư\right)}=\dfrac{3}{4}n_{Al}=\dfrac{3}{4}.0,4=0,3\left(mol\right)\\n_{Al_2O_3}=\dfrac{1}{2}n_{Al}=\dfrac{1}{2}.0,4=0,2\left(mol\right)\end{matrix}\right.\)

=> V_{O2 (dư)} = (0,6 - 0,3).22,4 = 6,72 (l)

c, m_Al2O3 = 0,2.102 = 20,4 (g)

\(n_{Al}=\dfrac{10,8}{17}=0,4\left(mol\right)\\ n_{O_2}=\dfrac{13,44}{22,4}=0,6\left(mol\right)\\ pthh:4Al+3O_2\underrightarrow{t^o}2Al_2O_3\) 
\(LTL:\dfrac{0,4}{4}< \dfrac{0,6}{3}\) 
=> O2 dư P hết 
\(n_{Al_2O_3}=\dfrac{1}{2}n_{Al}=0,2\left(mol\right)\\ m_{Al_2O_3}=0,2.102=20,4g\)

\(n_{Al}=\dfrac{10,8}{27}=0,4mol\)

\(n_{O_2}=\dfrac{13,44}{22,4}=0,6mol\)

    \(4Al+3O_2\rightarrow\left(t^o\right)2Al_2O_3\)

Xét: \(\dfrac{0,4}{4}\) < \(\dfrac{0,6}{3}\)                  ( mol )

         0,4     0,3            0,2   ( mol )

Chất dư là O₂

\(m_{O_2\left(dư\right)}=\left(0,6-0,3\right).32=9,6g\)

\(m_{Al_2O_3}=0,2.102=20,4g\)

2Cu+O2-to>2CuO

0,1-----0,05-----0,1

4P+5O2-to>2P2O5

n Cu=\(\dfrac{6,4}{64}\)=0,1 mol

=>VO2=0,05.22,4=1,12l

=>m CuO=0,1.80=8g

b) 

thiếu đề

a. \(n_{Cu}=\dfrac{6.4}{64}=0,1\left(mol\right)\)

PTHH : 2Cu + O₂ -> 2CuO

            0,1       0,05     0,1

\(V_{O_2}=0,05.22,4=1,12\left(l\right)\)

\(m_{CuO}=0,1.81=8,1\left(g\right)\)

b. Thiếu số mol P