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a,\(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right);n_{H_2SO_4}=0,45.15=6,75\left(mol\right)\)
PTHH: 2Al + 3H2SO4 → Al2(SO4)3 + 3H2
Mol: 0,2 0,3 0,1 0,3
Ta có: \(\dfrac{0,2}{2}< \dfrac{6,75}{3}\) ⇒ Al pứ hết, H2SO4 dư
\(n_{H_2SO_4dư}=6,75-0,3=6,45\left(mol\right)\)
\(V_{H_2}=0,3.22,4=6,72\left(l\right)\)
b,\(m_{Al_2\left(SO_4\right)_3}=0,1.342=34,2\left(g\right)\)
a,\(m_{H_2SO_4}=20\%.294=58,8\left(g\right)\Rightarrow n_{H_2SO_4}=\dfrac{58,8}{98}=0,6\left(mol\right)\)
PTHH: CuO + H2SO4 → CuSO4 + H2O
Mol: x x
PTHH: 2Al + 3H2SO4 → Al2(SO4)3 + 3H2
Mol: y 1,5y
Ta có: \(\left\{{}\begin{matrix}80x+27y=29,4\\x+1,5y=0,6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0,3\\y=0,2\end{matrix}\right.\)
\(\%m_{CuO}=\dfrac{0,3.80.100\%}{29,4}=81,63\%;\%m_{Al}=100-81,63=18,34\%\)
b,
PTHH: 2Al + 3H2SO4 → Al2(SO4)3 + 3H2
Mol: 0,2 0,3
\(\Rightarrow V_{H_2}=0,3.22,4=6,72\left(l\right)\)
Bài 1 :
$S + O_2 \xrightarrow{t^o} SO_2$
Ta thấy :
$n_S = \dfrac{6,4}{32} = 0,2(mol) > n_{O_2} = \dfrac{3,36}{22,4} = 0,15(mol)$ nên S dư
$n_{SO_2} = n_{O_2} = 0,15(mol)$
$V_{SO_2} = 0,15.22,4 = 3,36(lít)$
Bài 2 :
Gọi $n_{Al} = a(mol) ; n_{Fe} = b(mol) \Rightarrow 27a + 56b = 13,8(1)$
$4Al + 3O_2 \xrightarrow{t^o} 2Al_2O_3$
$3Fe + 2O_2 \xrightarrow{t^o} Fe_3O_4$
Suy ra :
$102.0,5a + \dfrac{b}{3}.232 = 21,8(2)$
Từ (1)(2) suy ra a = 0,2 ; b = 0,15
$\%m_{Al} = \dfrac{0,2.27}{13,8}.100\% = 39,13\%$
$\%m_{Fe} = 60,87\%$
Oxit axit :
\(FeO\) : sắt (II) oxit
\(CaO\) : canxi oxit
\(Na_2O\) : Natri oxit
Oxit axit :
\(SO_2\) : lưu huỳnh đioxit
\(N_2O_5\) : đi nito pentaoxit
Axit :
\(H_3PO_4\) : axit photphoric
\(HBr\) : axit bromhidric
\(HNO_3\) : axit nitric
\(H_2SO_3\) : axit sunfuric
\(H_2S\) : axit sunfuhidric
Bazo :
\(Cu\left(OH\right)_2\) : đồng (II) hidroxit
\(Ca\left(OH\right)_2\) : canxi hidroxit
\(Al\left(OH\right)_3\) : nhôm hidroxit
Muối :
\(Fe\left(NO_3\right)_3\) : muối sắt (III) nitrat
Chúc bạn học tốt
Oxit :
$FeO$ : Sắt II oxit
$SO_2$ : Lưu huỳnh đioxit
$N_2O_5$: đinito pentaoxit
$CaO$ : Canxi oxit
$Na_2O$ : Natri oxit
Axit :
$H_3PO_4$ : Axit photphoric
$HBr$ : Axit bromhidric
$HNO_3$: Axit nitric
$H_2SO_4$: Axit sunfuric
$H_2S$ : Axit sunfuhidric
Bazo :
$Cu(OH)_2$ : Đồng II hidroxit
$Ca(OH)_2$ : Canxi hidroxit
$Al(OH)_3$ : Nhôm hidroxit
Muối :
$Fe(NO_3)_3$ : Sắt III nitrat
Bài 1 :
\(n_{H2}=\dfrac{0,896}{22,4}=0,04\left(mol\right)\)
Pt : \(Zn+2HCl\rightarrow ZnCl_2+H_2|\)
1 2 1 1
0,04 0,08 0,04
a) \(n_{Zn}=\dfrac{0,04.1}{1}=0,04\left(mol\right)\)
⇒ \(m_{Zn}=0,04.65=2,6\left(g\right)\)
b) \(n_{HCl}=\dfrac{0,04.2}{1}=0,08\left(mol\right)\)
⇒ \(m_{HCl}=0,08.36,5=2,92\left(g\right)\)
\(C_{ddHCl}=\dfrac{2,92.100}{200}=1,46\)0/0
Chúc bạn học tốt
\(n=\dfrac{m}{M}\Leftrightarrow m=n.M\Leftrightarrow M=\dfrac{m}{n}\)
\(n=\dfrac{V}{22,4}\Leftrightarrow V=22,4.n\)
\(n=V_{dd}.C_M\Leftrightarrow C_M=\dfrac{n}{V_{dd}}\Leftrightarrow V_{dd}=\dfrac{n}{C_M}\)
\(C\%=\dfrac{m_{ct}.100\%}{m_{dd}}\Leftrightarrow m_{dd}=\dfrac{m_{ct}.100\%}{C\%}\Leftrightarrow m_{ct}=\dfrac{C\%.m_{dd}}{100\%}\)
n = \(\dfrac{m}{M}\) ⇔ m = n * M ⇔ M = \(\dfrac{m}{n}\)
n = \(\dfrac{V}{22,4}\) ⇒ V = n * 22,4
n = V(dd) * \(C_m\) ⇒ \(C_m\) = \(\dfrac{n}{V_{dd}}\) ⇒ \(V_{dd}\) = \(\dfrac{n}{C_m}\)
C% = \(\dfrac{m_{ct}}{m_{dd}}\)100% ⇔ \(m_{dd}\) = \(\dfrac{m_{ct}\cdot100\%}{C\%}\) ⇔ \(m_{ct}\) = \(\dfrac{m_{dd}\cdot C\%}{100\%}\)