a) PTHH: \(BaCO_3+2HCl\rightarrow BaCl_2+H_2O+CO_2\uparrow\)

Ta có: \(\left\{{}\begin{matrix}n_{BaCO_3}=\dfrac{68,95}{197}=0,35\left(mol\right)\\n_{HCl}=0,25\cdot3,2=0,8\left(mol\right)\end{matrix}\right.\)

Xét tỉ lệ: \(\dfrac{0,35}{1}< \dfrac{0,8}{2}\) \(\Rightarrow\) Axit còn dư 

\(\Rightarrow n_{HCl\left(dư\right)}=0,8-0,35\cdot2=0,1\left(mol\right)\) \(\Rightarrow m_{HCl\left(dư\right)}=0,1\cdot36,5=3,65\left(g\right)\)

b+c) Theo PTHH: \(\left\{{}\begin{matrix}n_{CO_2}=n_{BaCl_2}=0,35\left(mol\right)\\n_{HCl\left(dư\right)}=0,1\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}V_{CO_2}=0,35\cdot22,4=7,84\left(l\right)\\C_{M_{BaCl_2}}=\dfrac{0,35}{0,25}=1,4\left(M\right)\\C_{M_{HCl\left(dư\right)}}=\dfrac{0,1}{0,25}=0,4\left(M\right)\end{matrix}\right.\) 

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

nHCl = 0.4*1.5 = 0.6 (mol) 

\(Al_2O_3+6HCl\rightarrow2AlCl_3+3H_2O\)

nAlCl3 = 0.6*2/6 = 0.2 (mol) 

mAlCl3 = 0.2*133.5 = 26.7 (g) 

CM AlCl3 = 0.2/0.4 = 0.5 (M) 

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: \(n_{HCl}=2n_{CuO}+6n_{Fe_2O_3}=1\left(mol\right)\)

\(\Rightarrow V_{ddHCl}=\dfrac{1}{0,5}=2\left(l\right)\)

\(n_{Al}=\dfrac{0,54}{27}=0,02mol\\ n_{H_2SO_4}=0,07.0,5=0,035mol\\ 2Al+3H_2SO_4->Al_2\left(SO_4\right)_3+3H_2\\ \Rightarrow H_2SO_4:dư\\ V=\dfrac{3}{2}.0,02.22,4=0,672L\\ C_{M\left(H_2SO_4\right)}=\dfrac{0,005}{0,07}=0,071M\\ C_{M\left(Al_2\left(SO_4\right)_3\right)}=\dfrac{0,01}{0,07}=\dfrac{1}{7}\left(M\right)\\ CuO+H_2-t^{^0}->Cu+H_2O\\ n_{H_2}=0,03mol\\ n_{CuO}=\dfrac{6,4}{80}=0,08\Rightarrow CuO:dư\\ m_{rắn}=6,4-16.0,03=5,92g\)

CuO + 2Hcl = CuCl2 + H20

0.075...0,15...0,075

Cm = n/V = 0,15/0,4= 0,375

mCuCl2 = 0,075.135=10,125

a) CuO + 2HCl → CuCl₂ + H₂O

b)0,075.....0,15.......0,075

⇒C_M\(_{HCl}\)=\(\dfrac{1,15}{0,4}\)=0,375M

c)m\(_{CuCl_2}\)=0.075.135=10,125(g)

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 :

$C\%_{CuSO_4} = \dfrac{0,02.160}{101,6}.100\% = 3,15\%$

$C\%_{H_2SO_4\ dư} = \dfrac{100.20\% - 0,02.98}{101,6}.100\% = 17,6\%$

CÂU 2

m_H2SO4=100.20%=20(g)

n_H2SO4=20/98=0,2(mol)

n_CuO=1,6/80=0,02(mol)

PTHH : CuO+H₂SO₄-->CuSO₄+H₂O(1)

bài        0,02    0,2         0,02       0,02             (mol)

có:0,02/1<0,2/1---->CuO hết,H₂SO₄ dư

từ pt(1)-->n_CuSO4=0,02(mol)--->m_CuSO4=0,02.160=3,2(g)

khối lượng dd sau pư là:1,6+100-0,02.18=101,24(g)

-->C_%(CuSO4)=3,2/101,24.100%=3,16%

CÂU 3

m_HCl=10.21,9%=2,19(g)

-->n_HCl=2,19/36,5=0,06(mol)

gọi tên KL là M.M_M=M(g/mol)

PTHH: MO+2HCl-->MCl₂+H₂O(1)

          0,03     0,06                       (mol)

từ pt 1-->n_MO=0,03(mol)

--->M_MO=2,4/0,03=80(g/mol)

--->M=80-16=64(g/mol)

--->M là Cu