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\(n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)
PTHH: Zn + 2HCl ---> FeCl2 + H2
0,3<---------------0,3<----0,3
=> \(\left\{{}\begin{matrix}m=0,3.65=19,5\left(g\right)\\m_{muối}=0,3.136=40,8\left(g\right)\\V_{ddHCl}:thiếu.C_M\end{matrix}\right.\)
\(n_{Fe_2O_3}=\dfrac{32}{160}=0,2\left(mol\right)\)
PTHH: Fe2O3 + 3H2 --to--> 2Fe + 3H2O
LTL: \(0,2>\dfrac{0,3}{3}\) => Fe2O3 dư
Theo pthh: nFe2O3 (pư) = \(\dfrac{1}{3}n_{H_2}=\dfrac{1}{3}.0,3=0,1\left(mol\right)\)
nFe = \(\dfrac{2}{3}n_{H_2}=\dfrac{2}{3}.0,3=0,2\left(mol\right)\)
=> mchất rắn = 0,1.160 + 0,2.56 = 27,2 (g)
\(n_{H_2}=\dfrac{13,44}{22,4}=0,6mol\)
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
0,6 1,2 0,6 0,6 ( mol )
\(m_{Fe}=0,6.56=33,6g\)
\(m_{FeCl_2}=0,6.127=76,2g\)
\(C_{M_{HCl}}=\dfrac{1,2}{0,6}=2M\)
`Fe + 2HCl -> FeCl_2 + H_2↑`
`0,3` `0,6` `0,3` `0,3` `(mol)`
`n_[H_2] = [ 6,72 ] / [ 22,4 ] = 0,3 (mol)`
`-> m_[Fe] = 0,3 . 56 = 16,8 (g)`
`-> m_[FeCl_2] = 0,3 . 127 = 38,1 (g)`
`b) C_[M_[HCl]] = [ 0,6 ] / [ 0,3 ] = 2 (M)`
\(n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)
PTHH: Fe + 2HCl ---> FeCl2 + H2
0,3<---0,6<------0,3<-----0,3
=> \(\left\{{}\begin{matrix}m_{Fe}=0,3.56=16,8\left(g\right)\\m_{FeCl_2}=127.0,3=38,1\left(g\right)\\C_{M\left(HCl\right)}=\dfrac{0,6}{0,3}=2M\end{matrix}\right.\)
Ta có: \(n_{H_2}=\dfrac{1,12}{22,4}=0,05\left(mol\right)\)
a, PT: \(Fe+2HCl\rightarrow FeCl_2+H_2\)
_____0,05__0,1____________0,05 (mol)
b, mFe = 0,05.56 = 2,8 (g)
c, mHCl = 0,1.36,5 = 3,65 (g)
\(\Rightarrow m_{ddHCl}=\dfrac{3,65}{10\%}=36,5\left(g\right)\)
Bạn tham khảo nhé!
a) $n_{H_2SO_4} = \dfrac{44,1}{98} = 0,45(mol)$
$2Al + 3H_2SO_4 \to Al_2(SO_4)_3 + 3H_2$
Theo PTHH :
$n_{Al} = \dfrac{2}{3}n_{H_2SO_4} = 0,3(mol)$
$m_{Al} = 0,3.27 = 8,1(gam)$
b) $n_{H_2} = n_{H_2SO_4} = 0,45(mol)$
$\Rightarrow V_{H_2} = 0,45.22,4 =1 0,08(lít)$
c)
Cách 1 : $n_{Al_2(SO_4)_3} = \dfrac{1}{3}n_{H_2SO_4} = 0,15(mol)$
$\Rightarrow m_{Al_2(SO_4)_3} = 0,15.342 = 51,3(gam)$
Cách 2 : Bảo toàn khối lượng, $m_{Al_2(SO_4)_3} = 8,1 + 44,1 - 0,45.2 = 51,3(gam)$
PTHH: Zn + 2 HCl -> ZnCl2 + H2
nZn=13/65=0,2(mol)
nH2=nZnCl2=nZn=0,2(mol)
a) V(H2,đktc)=0,2.22,4=4,48(l)
b) nCuO= 24/80=0,3(mol)
PTHH: CuO + H2 -to-> Cu + H2O
Ta có: 0,3/1 > 0,2/1 => CuO dư, H2 hết, tính theo nH2
=> nCu=nH2=0,2(mol)
=> mCu=0,2.64=12,8(g)
c) 2 H2 + O2 -to-> H2O
nO2= 1/2 . nH2= 1/2 . 0,2=0,1(mol)
=> m(O2,đktc)=0,1.22,4=2,24(l)
a) PTHH: \(Fe+2HCl\rightarrow FeCl_2+H_2\uparrow\)
b+c) Ta có: \(n_{Fe}=\dfrac{8,4}{56}=0,15\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}n_{H_2}=0,15\left(mol\right)\\n_{HCl}=0,3\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}V_{H_2}=0,15\cdot22,4=3,36\left(l\right)\\m_{ddHCl}=\dfrac{0,3\cdot36,5}{10,95\%}=100\left(g\right)\end{matrix}\right.\)
d) PTHH: \(H_2+CuO\xrightarrow[]{t^o}Cu+H_2O\)
Ta có: \(\left\{{}\begin{matrix}n_{H_2}=0,15\left(mol\right)\\n_{CuO}=\dfrac{24}{80}=0,3\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\) CuO còn dư
\(\Rightarrow n_{CuO\left(dư\right)}=0,15\left(mol\right)\) \(\Rightarrow m_{CuO\left(dư\right)}=0,15\cdot80=12\left(g\right)\)
Ta có: \(n_{H_2SO_4}=\dfrac{58,8}{98}=0,6\left(mol\right)\)
\(PTHH:2Al+3H_2SO_4--->Al_2\left(SO_4\right)_3+3H_2\uparrow\)
0,4 <--- 0,6 -----------> 0,2 --> 0,6
\(\Rightarrow\left\{{}\begin{matrix}m_{Al}=0,4.27=10,8\left(g\right)\\m_{Al_2\left(SO_4\right)_3}=0,2.342=68,4\left(g\right)\\V_{H_2}=0,6.22,4=13,44\left(lít\right)\end{matrix}\right.\)
a) pt
1) Zn + 2HCl -> ZnCl2 + H2
2) H2 + CuO -> Cu + H2O
b) ta có mZn = 6,5 g
=> nZn = 0,1 mol
pt 1)
1) Zn + 2HCl -> ZnCl2 + H2
..0,1mol..0,1mol..0,1mol.....0,1mol
theo pt 1) ta có nZn = nH2 = 0,1mol
pt 2)
2) H2 + CuO -> Cu + H2O
...0,1mol...0,1mol...0,1mol..0,1mol
theo pt 2) ta có nH2 = nCu = 0,1mol
=> mCu = 0,1 . 64 = 6,4 g
1. PTHH: \(Zn+2HCl\rightarrow ZnCl_2+H_2\) (1)
\(H_2+CuO\rightarrow Cu+H_2O\) (2)
2. \(n_{Zn}=\dfrac{6,5}{65}=0.1mol\)
Từ phương trình (1) : \(n_{H_2}=n_{Zn}=0.1mol\)
Vì lượng \(H_2\) ở cả hai phương trình bằng nhau nên cùng bằng 0,1 mol.
Từ phương trình (2) :\(n_{Cu}=n_{H_2}=0,1mol\)
\(\rightarrow m_{Cu}=0,1.64=6,4g\)