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

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 \(m_{O_2(dư)}=1,056(g)\)

PTHH: \(4Fe+3O_2\rightarrow2Fe_2O_3\)

Ta có: \(\left\{{}\begin{matrix}n_{Fe}=\dfrac{16,8}{56}=0,3\left(mol\right)\\n_{Fe_2O_3}=\dfrac{16}{160}=0,1\left(mol\right)\end{matrix}\right.\)

Xét tỉ lệ: \(\dfrac{0,3}{4}>\dfrac{0,1}{2}\) \(\Rightarrow\) Sắt còn dư

\(\Rightarrow\left\{{}\begin{matrix}n_{O_2}=\dfrac{3}{2}n_{Fe_2O_3}=0,15\left(mol\right)\\n_{Fe\left(dư\right)}=0,1\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{Fe\left(dư\right)}=0,1\cdot56=5,6\left(g\right)\\V_{O_2}=0,15\cdot22,4=3,36\left(l\right)\end{matrix}\right.\)

1/  Fe +2 HCl --------> FeCl2 + H2

\(n_{Fe}=\dfrac{2,8}{56}=0,05\left(mol\right)\)

\(n_{H_2}=n_{Fe}=0,05\left(mol\right)\Rightarrow V_{H_2}=0,05.22,4=1,12\left(l\right)\)

\(n_{HCl}=2n_{Fe}=0,1\left(mol\right)\Rightarrow m_{HCl}=0,1.36,5=3,65\left(g\right)\)

Câu 6 : 

1) $n_{Fe} = \dfrac{2,8}{56} = 0,05(mol)$
Fe + 2HCl → FeCl₂ + H₂

0,05...0,1....................0,05......(mol)$
$V_{H_2} = 0,05.22,4 = 1,12(lít)$
$m_{HCl} = 0,1.36,5 = 3,65(gam)$

2)

a) $CH_4 + 2O_2 \xrightarrow{t^o} CO_2 + 2H_2O$

$V_{O_2} = 2V_{CH_4} = 4(lít)$

b) $n_{CO_2} = n_{CH_4} = 0,15(mol) \Rightarrow V_{CO_2} = 0,15.22,4 = 3,36(lít)$

c) $d_{CH_4/kk} = \dfrac{16}{29} = 0,552$

Vậy khí metan nhẹ hơn không khí 0,552 lần

1. Ta có: \(n_{Fe}=\dfrac{2,8}{56}=0,05\left(mol\right)\)

PT: \(Fe+2HCl\rightarrow FeCl_2+H_2\)

___0,05____0,1__________0,05 (mol)

\(\Rightarrow V_{H_2}=0,05.22,4=1,12\left(l\right)\)

\(m_{HCl}=0,1.36,5=3,65\left(g\right)\)

2. Ở cùng điều kiện nhiệt độ và áp suất, tỉ lệ số mol cũng là tỉ lệ thể tích.

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

_______2_____4 (lít)

⇒ V_O2 = 4 (l)

b, Theo PT: \(n_{CO_2}=n_{CH_4}=0,15\left(mol\right)\)

\(\Rightarrow V_{CO_2}=0,15.22,4=3,36\left(l\right)\)

c, M_CH4 = 16 (g/mol)

⇒ Khí metan nhẹ hơn không khí 16/29 lần.

Bạn tham khảo nhé!

\(n_{Fe}=\dfrac{12.6}{56}=0.225\left(mol\right)\)

\(n_{O_2}=\dfrac{4.2}{22.4}=0.1875\left(mol\right)\)

\(3Fe+2O_2\underrightarrow{^{^{t^0}}}Fe_3O_4\)

\(3.........2\)

\(0.225......0.1875\)

Lập tỉ lệ : \(\dfrac{0.225}{3}< \dfrac{0.1875}{2}\Rightarrow O_2dư\)

\(m_{O_2\left(dư\right)}=\left(0.1875-0.225\cdot\dfrac{2}{3}\right)\cdot32=1.2\left(g\right)\)

\(m_{Fe_3O_4}=\dfrac{0.225}{3}\cdot232=17.4\left(g\right)\)

\(n_{KMnO_4}=\dfrac{31,6}{158}=0,2mol\)

\(2KMnO_4\underrightarrow{t^o}K_2MnO_4+MnO_2+O_2\uparrow\)

0,2                 0,1              0,1        0,1

a)\(V_{O_2}=0,1\cdot22,4=2,24l\)

b)\(m_{CRắn}=m_{K_2MnO_4}+m_{MnO_2}=0,1\cdot197+0,1\cdot87=28,4g\)

c)\(n_{CH_4}=\dfrac{11,2}{22,4}=0,5mol\)

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

0,5        0,1    0              0

0,05      0,1    0,05         0,1

0,45       0      0,05         0,1

\(V_{CO_2}=0,05\cdot22,4=1,12l\)

\(m_{H_2O}=0,1\cdot18=1,8g\)

PTHH: 2KMnO4--t^o-> K2MnO4+MnO2+O2

              0,2----------------0,1---------0,1-----0,1

b, n_KMnO4= \(\dfrac{31,6}{158}\)=0,2 mol

Theo pt: n_O2=\(\dfrac{1}{2}\).0,2=0,1 mol

=> V_O2= 0,1.22,4= 2,24 l

=>m cr=0,1.197+0,1.87=28,4g

CH4+2O2-to>CO2+2H2O

            0,5-----0,25-----0,5

n CH4=\(\dfrac{11,2}{22,4}\)=0,5 mol

=>Oxi du

=>V CO2=0,25.22,4=5,6l

=>m H2O=0,5.18=9g

`a) PTHH:`

`2Fe + O_2` $\xrightarrow[]{t^o}$ `2FeO`

`0,5`    `0,25`        `0,5`                `(mol)`

`b) n_[Fe] = 28 / 56 = 0,5 (mol)`

`=> m_[FeO] = 0,5 . 72 = 36 (g)`

`c) V_[O_2] = 0,25 . 22,4 = 5,6 (l)`