Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(n_{Fe}=\dfrac{5.6}{56}=0.1\left(mol\right)\)
\(n_{H_2SO_4}=\dfrac{14.7}{98}=0.15\left(mol\right)\)
\(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)
\(1...........1\)
\(0.1............0.15\)
\(LTL:\dfrac{0.1}{1}< \dfrac{0.15}{1}\Rightarrow H_2SO_4dư\)
\(V_{H_2}=0.1\cdot22.4=2.24\left(l\right)\)
\(m_{FeSO_4}=0.1\cdot152=15.2\left(g\right)\)
\(n_{Fe}=0,1\left(mol\right)\); \(n_{H2SO4}=0,15\left(mol\right)\)
PTHH: \(Fe+H_2SO_4\rightarrow FeSO_4+H_2\uparrow\)
LTL: 0,1 < 0,15 (mol)
Pư: 0,1→ 0,1 → 0,1 → 0,1 (mol)
Sau pư: 0 : 0,05 (mol)
a) \(V_{H_2\left(đktc\right)}=n.22,4=0,1.22,4=2,24\left(l\right)\)
b) \(m_{FeSO_4}=n.M=0,1.152=15,2\left(g\right)\)
a. \(n_{Zn}=\dfrac{2,6}{65}=0,04\left(mol\right)\)
\(PTHH:Zn+H_2SO_4\rightarrow ZnSO_4+H_2\uparrow\)
- Mol theo PTHH : \(1:1:1:1\)
- Mol theo phản ứng : \(0,04\rightarrow0,04\rightarrow0,04\rightarrow0,04\)
\(\Rightarrow m_{ZnSO_4}=n_{ZnSO_4}.M_{ZnSO_4}=0,04.161=6,44\left(g\right)\)
b. Từ a. suy ra : \(V_{H_2}=n_{H_2}.22,4=0,04.22,4=0,896\left(l\right)\)
c. Từ a. suy ra : \(n_{H_2}=0,04\left(mol\right)\)
\(PTHH:H_2+PbO\underrightarrow{t^o}Pb+H_2O\)
- Mol theo PTHH : \(1:1:1:1\)
- Mol theo phản ứng : \(0,04\rightarrow0,04\rightarrow0,04\rightarrow0,04\)
\(\Rightarrow m_{Pb}=n_{Pb}.M_{Pb}=0,04.207=8,28\left(g\right)\)
`Mg + H_2 SO_4 -> MgSO_4 + H_2`
`0,45` `0,45` `0,45` `(mol)`
`n_[H_2 SO_4] = [ [ 14,7 ] / 100 . 300 ] / 98 = 0,45 (mol)`
`a) V_[H_2] = 0,45 . 22,4 = 10,08 (l)`
`b) m_[MgSO_4] = 0,45 . 120 = 54 (g)`
\(n_{H_2SO_4}=\dfrac{300.14,7\%}{98}=0,45\left(mol\right)\\
pthh:Mg+2HCl\rightarrow MgCl_2+H_2\)
0,45 0,225 0,225
\(V_{H_2}=0,225.22,4=5,04l\\
m_{MgCl_2}=95.0,225=21,375g\)
\(n_K=\dfrac{7,8}{39}=0,2\left(mol\right)\\ a,2K+2H_2O\rightarrow2KOH+H_2\uparrow\\ n_{H_2}=\dfrac{0,2}{2}=0,1\left(mol\right)\\ V_{H_2\left(\text{đ}ktc\right)}=0,1.22,4=2,24\left(l\right)\\ b,n_{KOH}=n_K=0,2\left(mol\right)\\ m_{KOH}=0,2.56=11,2\left(g\right)\\ c,m_{\text{dd}sau}=m_K+m_{H_2O}-m_{H_2}\)
Nhưng chưa có KL nước?
\(n_{Al}=\dfrac{2,7}{27}=0,1\left(mol\right)\\ n_{H_2SO_4}=\dfrac{39,2}{98}=0,4\left(mol\right)\\ PTHH:2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\\ LTL:\dfrac{0,1}{2}< \dfrac{0,4}{3}\rightarrow H_2SO_4\text{ dư}\)
Theo pthh:
\(n_{H_2}=\dfrac{3}{2}n_{Al}=\dfrac{3}{2}.0,1=0,15\left(mol\right)\\ \rightarrow V_{H_2}=0,15.22,4=3,36\left(l\right)\\ PTHH:Fe_2O_3+3H_2\underrightarrow{t^o}2Fe+3H_2O\\ Mol:0,05\leftarrow0,15\\ \rightarrow m_{Fe_2O_3}=0,05.160=8\left(g\right)\)
\(n_{Fe}=\dfrac{5.6}{56}=0.1\left(mol\right)\)
\(n_{HCl}=\dfrac{10.95}{36.5}=0.3\left(mol\right)\)
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
Lập tỉ lệ :
\(\dfrac{0.2}{1}>\dfrac{0.3}{2}\Rightarrow Fedư\)
Khi đó :
\(n_{FeCl_2}=n_{H_2}=\dfrac{1}{2}\cdot n_{HCl}=\dfrac{1}{2}\cdot0.3=0.15\left(mol\right)\)
\(m_{FeCl_2}=0.15\cdot127=19.05\left(g\right)\)
\(V_{H_2}=0.15\cdot22.4=3.36\left(l\right)\)
\(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\\ n_{HCl}=\dfrac{10,95}{36,5}=0,3\left(mol\right)\\ PTHH:Fe+2HCl\rightarrow FeCl_2+H_2\\ Vì:\dfrac{0,1}{1}< \dfrac{0,3}{2}\\ \Rightarrow HCldư\\ \Rightarrow n_{FeCl_2}=n_{Fe}=n_{H_2}=0,1\left(mol\right)\\ \Rightarrow m_{FeCl_2}=127.0,1=12,7\left(g\right)\\ V_{H_2\left(đktc\right)}=0,1.22,4=2,24\left(l\right)\)
nH2SO4= 0,125(mol)
nFe=0,2(mol)
PTHH: Fe + H2SO4 -> FeSO4 + H2
Vì: 0,125/1 < 0,2/1
=> Fe dư, H2SO4 hết, tính theo nH2SO4
-> nH2=nH2SO4=0,125(mol)
=>V(H2,đktc)=0,125.22,4=2,8(l)
a)\(n_{Al}=\dfrac{18}{27}=\dfrac{2}{3}mol\)
\(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)
\(\dfrac{2}{3}\) 1
\(V_{H_2}=1\cdot22,4=22,4l\)
b)\(n_{Fe_2O_3}=\dfrac{32}{160}=0,2mol\)
\(Fe_2O_3+3H_2\rightarrow2Fe+3H_2O\)
0,2 1 0,4 0,6
\(m_{Fe}=0,4\cdot56=22,4g\)
c)\(3Fe+2O_2\underrightarrow{t^o}Fe_3O_4\)
0,4 \(\dfrac{4}{15}\)
\(V_{O_2}=\dfrac{4}{16}\cdot22,4=\dfrac{448}{75}l\)
\(V_{kk}=5V_{O_2}=\dfrac{448}{15}l\approx29,87l\)
\(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\)
\(n_{H_2SO_4}=\dfrac{14,7}{98}=0,15\left(mol\right)\)
PT: Fe + H2SO4 → FeSO4 + H2
Trước 0,1 0,15 0 0 mol
Trong 0,1 0,1 0,1 0,1 mol
Sau 0 0,05 0,1 0,1 mol
a) \(V_{H_2\left(đktc\right)}=0,1.22,4=2,24\left(l\right)\)
b) \(m_{FeSO_4}=0,1.152=15,2\left(g\right)\)