a.b.c.\(n_{Mg}=\dfrac{m_{Mg}}{M_{Mg}}\dfrac{4,8}{24}=0,2mol\)

\(2Mg+O_2\rightarrow\left(t^o\right)2MgO\)

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

\(V_{O_2}=n_{O_2}.22,4=0,1.22,4=2,24l\)

\(m_{MgO}=n_{MgO}.M_{MgO}=0,2.40=8g\)

d. Sửa đề: tính khối lượng KMnO4

\(2KMnO_4\rightarrow\left(t^o\right)K_2MnO_4+MnO_2+O_2\)

    0,2                                                      0,1    ( mol )

\(m_{KMnO_4}=n_{KMnO_4}.M_{KMnO_4}=0,2.158=31,6g\)

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

Gộp cả phần a và b

Ta có: \(n_{Mg}=\dfrac{12}{24}=0,5\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}n_{O_2}=0,25mol\\n_{MgO}=0,5mol\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{O_2}=0,25\cdot22,4=5,6\left(l\right)\\m_{MgO}=0,5\cdot40=20\left(g\right)\end{matrix}\right.\)

a, Mg+O2​to-->2MgO

b,

nMg​=0,5(mol)

nO2=0,25 mol;nMgO=0,5mol

→;nMgO​=0,5(mol)

→VO2​=5,6(l)

a)\(n_{Cu}=\dfrac{3,2}{64}=0,05\left(m\right)\)

\(PTHH:2Cu+O_2\underrightarrow{t^O}2CuO\)

tỉ lệ       : 2         1       2

số mol  :0,05    0,025  0,05

\(m_{CuO}=0,05.80=4\left(g\right)\)

b)\(V_{O_2}=0,025.22,4=0,56\left(l\right)\)

\(V_{kk}=\dfrac{0,56}{20\%}=2,8\left(l\right)\)

c)\(n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(m\right)\)

\(PTHH:CuO+H_2\xrightarrow[]{}Cu+H_2O\)

tỉ lệ        :1          1         1       1

số mlo   :0,1       0,1       0,1     0,1

\(m_{Cu}=0,1.64=6,4\left(g\right)\)

a) nFe = 16,8/56 = 0,3 (mol)

PTHH: 3Fe + 2O2 -> (t°) Fe3O4

Mol: 0,3 ---> 0,2 ---> 0,1

mFe3O4 = 0,1 . 232 = 23,2 (g)

b) VO2 = 0,2 . 22,4 = 4,48 (l)

Vkk = 4,48 . 5 = 22,4 (l)

c) H = 100% - 20% = 80%

nO2 (LT) = 0,2 : 80% = 0,25 (mol)

PTHH: 2KMnO4 -> (t°) K2MnO4 + MnO2 + O2

nKMnO4 = 0,25 . 2 = 0,5 (mol)

mKMnO4 = 0,5 . 158 = 79 (g)

\(n_{Mg}=\dfrac{4.8}{24}=0.2\left(mol\right)\)

\(2Mg+O_2\underrightarrow{t^0}2MgO\)

\(0.2.......0.1........0.2\)

\(V_{O_2}=0.1\cdot22.4=2.24\left(l\right)\)

\(m_{MgO}=0.2\cdot40=8\left(g\right)\)

a) 

2Mg + O2 --t^o--> 2MgO

0,2----->0,1------>0,2     (mol)

n_Mg = 4,48/24 = 0,2 (mol)

=> V_O2 = 0,1.22,4 = 2,24 (lít)

b) m_MgO = 0,2.(24 + 16) = 8 (g) 

a) $2Mg + O_2 \xrightarrow{t^o} 2MgO$

Theo PTHH : n MgO = n Mg = 2,4/24 = 0,1(mol)

=> m MgO = 0,1.40 = 4(gam)

b) n O2 = 1/2 n Mg = 0,05(mol) 

=> V O2 = 0,05.22,4 = 1,12(lít)

=> V không khí = 5V O2 = 1,12.5 = 5,6(lít)

c) $Mg + 2HCl \to MgCl_2 + H_2$

n HCl = 200.15%/36,5 = 60/73(mol)

Ta thấy :

n Mg / 1 = 0,1 < n HCl / 2 = 30/73 suy ra HCl dư

n H2 = n Mg = 0,1(mol)

=> m dd sau pư = 2,4 + 200 - 0,1.2 = 202,2(gam)

Vậy :

C% MgCl2 = 0,1.95/202,2  .100% = 4,7%

C% HCl = (60/73 - 0,1.2).36,5/202,2  .100% = 11,23%

nK=0,2(mol)

PTHH: 4K + O2 -to-> 2 K2O

nK2O= 0,1(mol) => mK2O=0,1.94=9,4(g)

nO2=0,05(mol) -> V(O2,đktc)=0,05.22,4=1,12(l)

V(kk,dktc)=5.V(O2,dktc)=5.1,12=5,6(l)

Gọi $n_{Mg} = a(mol) ; n_{Al} = b(mol) \Rightarrow 24a + 27b = 10,35(1)$

$2Mg + O_2  \xrightarrow{t^o} 2MgO$

$4Al + 3O_2  \xrightarrow{t^o} 2Al_2O_3$

$n_{O_2} = \dfrac{1}{2}a + \dfrac{3}{4}b = \dfrac{5,88}{22,4} = 0,2625(2)$

Từ (1)(2) suy ra a = 0,15 ; b = 0,25

$m_{Mg} = 0,15.24 = 3,6(gam)$
$m_{Al} = 0,25.27 = 6,75(gam)$

a. \(n_P=\frac{6,2}{31}=0,2mol\)

\(V_{O_2}=V_{kk}.\frac{1}{5}=\frac{18,48}{5}=3,696l\)

\(n_{O_2}=\frac{3,696}{22,4}=0,165mol\)

PTHH: \(4P+5O_2\xrightarrow{t^o}2P_2O_5\)

Tỷ lệ \(\frac{0,2}{4}>\frac{0,165}{5}\)

Vậy P dư

\(n_{P\left(\text{phản ứng }\right)}=\frac{4}{5}n_{O_2}=0,132mol\)

\(n_{P\left(dư\right)}=0,2-0,132=0,068mol\)

\(\rightarrow m_{P\left(dư\right)}=0,068.31=2,108g\)

b. \(n_{P_2O_5}=\frac{2}{5}n_{O_2}=0,066mol\)

\(\rightarrow m_{P_2O_5}=0,066.142=9,372g\)

c. PTHH: \(2KClO_3\xrightarrow{t^o}2KCl+3O_2\)

\(n_{KClO_3}=\frac{2}{3}n_{O_2}=0,11mol\)

\(\rightarrow m_{KClO_3}=0,11.122,5=13,475g\)