\(n_{HCl}=1\cdot0,2=0,2\left(mol\right)\\ PTHH:MgO+2HCl\rightarrow MgCl_2+H_2O\\ a,n_{MgO}=\dfrac{1}{2}n_{HCl}=0,1\left(mol\right)\\ \Rightarrow m=m_{MgO}=0,1\cdot40=4\left(g\right)\\ b,n_{MgCl_2}=n_{MgO}=0,1\left(mol\right)\\ \Rightarrow m_{MgCl_2}=0,1\cdot95=9,5\left(g\right)\\ c,m_{CT_{HCl}}=0,2\cdot36,5=7,3\left(g\right)\\ \Rightarrow C\%_{HCl}=\dfrac{7,3}{250}\cdot100\%=2,92\%\)

\(n_{H_2O}=n_{MgO}=0,1\left(mol\right)\\ \Rightarrow m_{H_2O}=0,1\cdot18=1,8\left(g\right)\\ \Rightarrow m_{dd_{MgCl_2}}=4+250-1,8=252,2\left(g\right)\\ \Rightarrow C\%_{MgCl_2}=\dfrac{9,5}{252,2}\cdot100\%\approx3,77\%\)

Bài 1:

Ta có: \(n_{HCl}=0,2.0,5=0,1\left(mol\right)\)

BTNT H, có: \(n_{HCl}=2n_{H_2O}\Rightarrow n_{H_2O}=0,05\left(mol\right)\)

Theo ĐL BTKL, có: m _oxit + m_HCl = m_muối + m_H2O

⇒ m_muối = 2,8 + 0,1.36,5 - 0,05.18 = 5,55 (g)

Bài 2:

\(m_{KOH}=200.5,6\%=11,2\left(g\right)\Rightarrow n_{KOH}=\dfrac{11,2}{56}=0,2\left(mol\right)\)

PT: \(2KOH+CuCl_2\rightarrow2KCl+Cu\left(OH\right)_2\)

Theo PT: \(n_{Cu\left(OH\right)_2}=\dfrac{1}{2}n_{KOH}=0,1\left(mol\right)\)

\(\Rightarrow m_{Cu\left(OH\right)_2}=0,1.98=9,8\left(g\right)\)

  nH2O=0.2 
nCuO=x,nAl2O3=y,nFeO=z 
80x + 102y + 72z = 17.86 
x + z =0.2 
135x + 267y + 127z = 33.81 
=> y=0.03 => mAl2O3=3.06g =>D

a) \(n_{SO_3}=\dfrac{3,2}{80}=0,04\left(mol\right)\)

PTHH: SO₃ + H₂O --> H₂SO₄

          0,04------------->0,04

=> \(m_{H_2SO_4}=0,04.98=3,92\left(g\right)\)

b) \(n_{Na}=\dfrac{0,69}{23}=0,03\left(mol\right)\)

PTHH: 2Na + 2H₂O --> 2NaOH + H₂

            0,03------------>0,03

           2NaOH + H₂SO₄ --> Na₂SO₄ + 2H₂O

Xét tỉ lệ: \(\dfrac{0,03}{2}< \dfrac{0,04}{1}\)=> NaOH hết, H₂SO₄ dư

2NaOH + H₂SO₄ --> Na₂SO₄ + 2H₂O

0,03------>0,015---->0,015

\(\left\{{}\begin{matrix}n_{Na_2SO_4}=0,015\left(mol\right)\\n_{H_2SO_4\left(dư\right)}=0,025\left(mol\right)\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}m_{Na_2SO_4}=0,015.142=2,13\left(g\right)\\m_{H_2SO_4}=0,025.98=2,45\left(g\right)\end{matrix}\right.\)

c) \(n_{Na}=\dfrac{2,07}{23}=0,09\left(mol\right)\)

PTHH: 2Na + 2H₂O --> 2NaOH + H₂

          0,09-------------->0,09

Xét tỉ lệ: \(\dfrac{0,09}{2}>\dfrac{0,04}{1}\) => NaOH dư, H₂SO₄ hết

2NaOH + H₂SO₄ --> Na₂SO₄ + 2H₂O

0,08<-----0,04------>0,04

=> \(\left\{{}\begin{matrix}n_{NaOH\left(dư\right)}=0,01\left(mol\right)\\n_{Na_2SO_4}=0,04\left(mol\right)\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}m_{NaOH\left(dư\right)}=0,01.40=0,4\left(g\right)\\m_{Na_2SO_4}=0,04.142=5,68\left(g\right)\end{matrix}\right.\)

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) 

b)  

c)  

d) Bạn xem lại đề !

a) n Fe = 28/56 = 0,5(mol)

$Fe + 2HCl \to FeCl_2 + H_2$

Theo PTHH :

n HCl = 2n Fe = 1(mol)

=> m dd HCl = 1.36,5/10% = 365(gam)

b)

n FeCl2 = n H2 = n Fe = 0,5(mol)

Suy ra : 

V H2 = 0,5.22,4 = 11,2(lít)

m FeCl2 = 0,5.127 = 63,5(gam)

c)

Sau phản ứng: 

mdd = m Fe + mdd HCl - m H2 = 28 + 365 - 0,5.2 = 392(gam)

=> C% FeCl2 = 63,5/392   .100% = 16,2%

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câu a

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