a) n HCl = 0,2.2,5 = 0,5 mol

b) m HCl =200.7,3% = 14,6 gam

n HCl = 14,6/36,5 = 0,4 mol

c) m NaOH = 300.40% = 120 gam

n NaOH = 120/40 = 3(mol)

d) n NaOH = 0,5.0,5 = 0,25 mol

a)

$Fe + 2HCl \to FeCl_2 + H_2$
$FeO +2 HCl \to FeCl_2 + H_2O$

b)

Theo PTHH : 

$n_{Fe} = n_{H_2} = \dfrac{4,48}{22,4} = 0,2(mol)$

$\%m_{Fe} = \dfrac{0,2.56}{20}.100\% = 56\%$

$\%m_{FeO} = 100\% - 56\% = 44\%$

c) $n_{FeO} = \dfrac{11}{90}(mol)$
$n_{HCl} = 2n_{Fe} + 2n_{FeO} = \dfrac{29}{45}(mol)$

$m_{dd\ HCl} = \dfrac{ \dfrac{29}{45}.36,5}{7,3\%} = 322,22(gam)$

Bài 14 : 

\(a) n_{CuO} = \dfrac{8}{80} = 0,1(mol)\\ CuO + 2HCl \to CuCl_2 + H_2O\\ n_{HCl} = 2n_{CuO} = 0,2(mol)\\ m_{dd\ HCl} = \dfrac{0,2.36,5}{7,3\%} = 100(gam)\\ b) \text{Chất tan : } CuCl_2\\ n_{CuCl_2} = n_{CuO} = 0,1(mol)\\ m_{CuCl_2} = 0,1.135 = 13,5(gam)\)

Bài 15 : 

\(a) n_{Fe_2O_3} =\dfrac{4,8}{160} = 0,03(mol)\\ Fe_2O_3 + 3H_2SO_4 \to Fe_2(SO_4)_3 + 3H_2O\\ n_{H_2SO_4} = 3n_{Fe_2O_3} = 0,09(mol)\\ m_{dd\ H_2SO_4} = \dfrac{0,09.98}{9,8\%} = 90(gam)\\ b) \text{Chất tan : } Fe_2(SO_4)_3\\ n_{Fe_2(SO_4)_3} = n_{Fe_2O_3} = 0,03(mol)\\ m_{Fe_2(SO_4)_3} = 0,03.400 = 12(gam)\)

Bài 16 :

\(a) Fe + 2HCl \to FeCl_2 + H_2\\ n_{H_2} = n_{Fe} = \dfrac{11,2}{56} = 0,2(mol)\\ V_{H_2} = 0,2.22,4 = 4,48(lít)\\ b) n_{HCl} = 2n_{Fe} = 0,4(mol)\\ C\%_{HCl} = \dfrac{0,4.36,5}{200}.100\% = 7,3\%\)

\(n_{Na_2SO_3}=\dfrac{25,2}{126}=0,2(mol)\\ n_{HCl}=\dfrac{250.7,3\%}{100\%.36,5}=0,5(mol)\\ PTHH:Na_2SO_3+2HCl\to 2NaCl+H_2O+SO_2\uparrow\)

Vì \(\dfrac{n_{Na_2SO_3}}{1}<\dfrac{n_{HCl}}{2}\) nên \(HCl\) dư

\(a,n_{SO_2}=n_{Na_2SO_3}=0,2(mol)\\ \Rightarrow V_{SO_2}=0,2.22,4=4,48(l)\)

\(b,\) Chất tan trong dd sau phản ứng là \(NaCl\)

\(c,n_{NaCl}=2n_{Na_2SO_3}=0,4(mol)\\ \Rightarrow m_{NaCl}=0,4.58,5=23,4(g)\)

Câu trên mình sai nha

\(n_{Na_2SO_3}=\dfrac{25,2}{126}=0,2(mol)\\ n_{HCl}=\dfrac{250.7,3\%}{100\%.36,5}=0,5(mol)\\ PTHH:Na_2SO_3+2HCl\to 2NaCl+H_2O+SO_2\uparrow\)

Vì \(\dfrac{n_{Na_2SO_3}}{1}<\dfrac{n_{HCl}}{2}\) nên \(HCl\) dư

\(a,n_{SO_2}=n_{Na_2SO_3}=0,2(mol)\\ \Rightarrow V_{SO_2}=0,2.22,4=4,48(l)\)

\(b,\) Chất tan trong dd sau phản ứng gồm \(HCl\) dư và \(NaCl\)

\(c,n_{HCl(dư)}=0,5-0,2.2=0,1(mol);n_{NaCl}=0,4(mol)\\ \Rightarrow m_{\text{dd sau p/ứ}}=m_{HCl(dư)}+m_{NaCl}=0,1.36,5+0,4.58,5=27,05(g)\)

Na2SO3+2HCl->2NaCl+H2O+SO2

0,201-------0,402------0,402---------------0.201

n Na2SO3=0,201 mol

m HCl=18,25 g

->n HCl=0,5 mol

=>HCl dư

=>VSO2=0,201.22,4=4,5024l

b)

mNaCl=\(\dfrac{0,402.58,5}{25,4+250-0,201.64}\).100=8,95%

m Hcl dư=\(\dfrac{0,098.36,5}{25,4+250-0,201.64}.100=1,36\%\)

a) Gọi số mol Zn, Al là a,b (mol)

=> 65a + 27b = 9,2 (1)

\(n_{H_2}=\dfrac{0,5}{2}=0,25\left(mol\right)\)

PTHH: Zn + 2HCl --> ZnCl₂ + H₂

             a---->2a-------------->a

            2Al + 6HCl --> 2AlCl₃ + 3H₂

            b----->3b--------------->1,5b

=> a + 1,5b = 0,25 (2)

(1)(2) => a = 0,1 (mol); b = 0,1 (mol)

=> \(\left\{{}\begin{matrix}\%m_{Zn}=\dfrac{0,1.65}{9,2}.100\%=70,65\%\\\%m_{Al}=\dfrac{0,1.27}{9,2}.100\%=29,35\%\end{matrix}\right.\)

b) n_HCl = 2a + 3b = 0,5 (mol)

=> \(C\%_{dd.HCl}=\dfrac{0,5.36,5}{200}.100\%=9,125\%\)

a) \(m_{HCl}=200\cdot7,3\%=14,6\left(g\right)\)

b) \(n_{NaOH}=0,5\cdot1=0,5\left(mol\right)\) \(\Rightarrow m_{NaOH}=0,5\cdot40=20\left(g\right)\)

c) \(n_{CuSO_4}=0,2\cdot1,5=0,3\left(mol\right)\) \(\Rightarrow m_{CuSO_4}=0,3\cdot160=48\left(g\right)\)

d) Bạn xem lại đề !

a) mHCl=200⋅7,3%=14,6(g)mHCl=200⋅7,3%=14,6(g)

b) nNaOH=0,5⋅1=0,5(mol)nNaOH=0,5⋅1=0,5(mol) ⇒mNaOH=0,5⋅40=20(g)⇒mNaOH=0,5⋅40=20(g)

c) nCuSO4=0,2⋅1,5=0,3(mol)nCuSO4=0,2⋅1,5=0,3(mol) ⇒mCuSO4=0,3⋅160=48(g)⇒mCuSO4=0,3⋅160=48(g)

d) Bạn xem lại đề !

a)Gọi : \(\left\{{}\begin{matrix}n_{Mg}=a\left(mol\right)\\n_{MgO}=b\left(mol\right)\end{matrix}\right.\)⇒ 24a + 40b = 8,8(1)

\(Mg + 2HCl \to MgCl_2 + H_2\\ MgO + 2HCl \to MgCl_2 + H_2O\)

Theo PTHH : 

\(n_{MgCl_2} = a + b = \dfrac{28,5}{95} = 0,3(2)\)

Từ (1)(2) suy ra:  a = 0,2 ; b = 0,1

Vậy : 

\(m_{Mg} = 0,2.24 = 4,8(gam) ; m_{MgO} = 0,1.40 = 4(gam)\\ \%m_{Mg} = \dfrac{4,8}{8,8}.100\% = 54,54\%\\ \%m_{MgO} = 100\% -54,54\% = 45,45\%\)

b)

\(n_{HCl} = 2n_{MgCl_2} = 0,3.2 = 0,6(mol)\\ C\%_{HCl} = \dfrac{0,6.36,5}{200}.100\% = 10,95\%\)