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a)
$Al_2O_3 + 6HCl \to 2AlCl_3 + 3H_2O$
b)
n HCl = 0,4.1,5 = 0,6(mol)
n Al2O3 = 1/6 n HCl = 0,1(mol) => m = 0,1.102 = 10,2(gam)
n AlCl3 = 1/3 n HCl = 0,2(mol) => CM AlCl3 = 0,2/0,4 = 0,5M
a, \(CuO+2HCl\rightarrow CuCl_2+H_2O\)
\(Fe_2O_3+6HCl\rightarrow2FeCl_3+3H_2O\)
b, Gọi: \(\left\{{}\begin{matrix}n_{CuO}=x\left(mol\right)\\n_{Fe_2O_3}=y\left(mol\right)\end{matrix}\right.\) ⇒ 80x + 160y = 32 (1)
Theo PT: \(\left\{{}\begin{matrix}n_{CuCl_2}=n_{Cu}=x\left(mol\right)\\n_{FeCl_3}=2n_{Fe_2O_3}=2y\left(mol\right)\end{matrix}\right.\) ⇒ 135x + 325y = 59,5 (2)
Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}x=0,2\left(mol\right)\\y=0,1\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}m_{CuO}=0,2.80=16\left(g\right)\\m_{Fe_2O_3}=0,1.160=16\left(g\right)\end{matrix}\right.\)
c, Theo PT: \(n_{HCl}=2n_{CuO}+6n_{Fe_2O_3}=1\left(mol\right)\)
\(\Rightarrow V_{ddHCl}=\dfrac{1}{0,5}=2\left(l\right)\)
nHCl=0,3.2=0,6(mol)
a) PTHH: CuO +2 HCl -> CuCl2 + H2O
0,3_______________0,6___0,3(mol)
b) mCuO=0,3.80=24(g)
c) VddCuCl2=VddHCl=0,3(l)
=>CMddCuCl2=0,3/0,3=1(M)
d) m(muối)=0,3.135=40,5(g)
Câu 3 :
\(n_{HCl}=\dfrac{10\cdot21.9\%}{36.5}=0.06\left(mol\right)\)
\(AO+2HCl\rightarrow ACl_2+H_2O\)
\(0.03........0.06\)
\(M=\dfrac{2.4}{0.03}=80\left(\dfrac{g}{mol}\right)\)
\(\Rightarrow A=64\)
\(CuO\)
Câu 2 :
$n_{CuO} = \dfrac{1,6}{80} = 0,02(mol)$
$n_{H_2SO_4} = \dfrac{100.20\%}{98} = \dfrac{10}{49}$
$CuO + H_2SO_4 \to CuSO_4 + H_2O$
$n_{CuO} < n_{H_2SO_4}$ nên $H_2SO_4 dư
Theo PTHH :
$n_{CuSO_4} = n_{H_2SO_4\ pư} = n_{CuO} = 0,02(mol)$
$m_{dd} = 1,6 + 100 = 101,6(gam)$
Vậy :
$C\%_{CuSO_4} = \dfrac{0,02.160}{101,6}.100\% = 3,15\%$
$C\%_{H_2SO_4\ dư} = \dfrac{100.20\% - 0,02.98}{101,6}.100\% = 17,6\%$
a/ \(n_{KOH}=0,2.1=0,2\left(mol\right);n_{H_2SO_4}=0,3.1=0,3\left(mol\right)\)
PTHH: 2KOH + H2SO4 → K2SO4 + 2H2O
Mol: 0,2 0,1 0,1
Ta có: \(\dfrac{0,2}{2}< \dfrac{0,3}{1}\) ⇒ KOH hết, H2SO4 dư
b/ \(m_{H_2SO_4dư}=\left(0,3-0,1\right).98=19,6\left(g\right)\)
c/ Vdd sau pứ = 0,2 + 0,3 = 0,5 (l)
d/ \(C_{M_{ddK_2SO_4}}=\dfrac{0,1}{0,5}=0,2M\)
\(C_{M_{ddH_2SO_4dư}}=\dfrac{0,3-0,1}{0,5}=0,4M\)
a)
$n_{Fe} = \dfrac{11,2}{56} = 0,2(mol) ; n_{HCl} = 0,1.2 = 0,2(mol)$
$Fe + 2HCl \to FeCl_2 + H_2$
Ta thấy :
$n_{Fe} : 1 > n_{HCl} : 2$ nên Fe dư
$n_{H_2} = \dfrac{1}{2}n_{HCl} = 0,1(mol)$
$V_{H_2} = 0,1.22,4 = 2,24(lít)$
b)
$n_{Fe\ pư} = n_{H_2} = 0,1(mol)$
$\Rightarrow m_{Fe\ dư} = 11,2 - 0,1.56 = 5,6(gam)$
c)
$n_{FeCl_2} = n_{Fe\ pư} = 0,1(mol)$
$C_{M_{FeCl_2}} = \dfrac{0,1}{0,1} = 1M$
