Cho kim loại Magie tác dụng vừa đủ với 200 gam dung dịch axit axetic 15%.

a. Tính khối lượng Magie phản ứng ?

b. Tính nồng độ phần trăm dung dịch muối thu được sau phản ứng ?

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mCH3COOH= 200.15%= 30(g) => nCH3COOH= 30/60=0,5(mol)

a) Mg + 2 CH3COOH -> (CH3COO)2Mg + H2

0,25___0,5_______0,25_____________0,25(mol)

mMg= 0,25.24= 6(g)

b) m(CH3COO)2Mg=142.0,25=35,5(g)

mdd(CH3COO)2Mg= 6+200-0,25.2=205,5(g)

=> \(C\%dd\left(CH3COO\right)2Mg=\frac{35,5}{205,5}.100\approx17,275\%\)

\(n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\\ a.Mg+2HCl\rightarrow MgCl_2+H_2\\ n_{H_2}=n_{MgCl_2}=n_{Mg}=0,2\left(mol\right)\\ V_{H_2\left(đktc\right)}=0,2.22,4=4,48\left(l\right)\\ b.n_{HCl}=0,2.2=0,4\left(mol\right)\\ m_{ddHCl}=\dfrac{0,4.36,5.100}{7,3}=200\left(g\right)\\ c.m_{ddsau}=4,8+200-0,2.2=204,4\left(g\right)\\ C\%_{ddMgCl_2}=\dfrac{0,2.95}{204,4}.100\approx9,295\%\\ d.V_{ddHCl}=\dfrac{200}{1,05}=\dfrac{4000}{21}\left(ml\right)=\dfrac{4}{21}\left(l\right)\\ C_{MddHCl}=\dfrac{0,4}{\dfrac{4}{21}}=2,1\left(M\right)\)

a) $n_{Fe} = \dfrac{11,2}{56} = 0,2(mol)$
$Fe + 2HCl \to FeCl_2 + H_2$
$n_{HCl} =2 n_{Fe} = 0,2.2 = 0,4(mol)$
$C\%_{HCl} = \dfrac{0,4.36,5}{200}.100\% = 7,3\%$

b) $n_{H_2} = n_{FeCl_2} = n_{Fe} = 0,2(mol)

Sau phản ứng, $m_{dd} = 11,2 + 200 - 0,2.2 = 210,8(gam)$
$C\%_{FeCl_2} = \dfrac{0,2.127}{210,8}.100\% = 12,05\%$

`a)PTHH:`

`Mg + H_2 SO_4 -> MgSO_4 + H_2`

`0,2`                                   `0,2`      `0,2`        `(mol)`

`n_[Mg]=[4,8]/24=0,2(mol)`

`b)m_[MgSO_4]=0,2.120=24(g)`

`c)C%_[MgSO_4]=24/[4,8+50-0,2.2].100~~44,12%`

100.\(\approx\)

nMg=9.6/24=0.4mol

Mg + 2HCl -> MgCl2 + H2

(mol) 0.4 0.4 0.4

mMgCl2=0.4*95=38g

mdd=mMg + mddHCl - mH2= 9.6+120-0.4*2=128.8g

C%MgCl2=38/128.8*100=29.5%

a)

Gọi $n_{Zn} = a(mol) ; n_{Al} = b(mol) \Rightarrow 65a + 27b = 11,9(1)$

$Zn + 2HCl \to ZnCl_2 + H_2$
$2Al + 6HCl \to 2AlCl_3 + 3H_2$
Theo PTHH : 

$n_{H_2} = a + 1,5b = \dfrac{8,96}{22,4} = 0,4(2)$

Từ (1)(2) suy ra : a = 0,1; b = 0,2

$m_{Zn} = 0,1.65 = 6,5(gam)$

$m_{Al} = 0,2.27 = 5,4(gam)$

b) $n_{HCl} = 2n_{H_2} = 0,8(mol)$
$C\%_{HCl} = \dfrac{0,8.36,5}{125}.100\% = 23,36\%$

Câu 2:

\(n_{MgBr_2}=\dfrac{14,72}{184}=0,08\left(mol\right)\\ Mg+Br_2\rightarrow MgBr_2\\ n_{Mg}=n_{Br_2}=n_{MgBr_2}=0,08\left(mol\right)\\ a=m_{Mg}=24.0,08=1,92\left(g\right)\\ m_{Br_2}=160.0,08=12,8\left(g\right)\)

Câu 1:

\(n_{AlBr_3}=\dfrac{106,8}{267}=0,4\left(mol\right)\\ 2Al+3Br_2\rightarrow2AlBr_3\\ n_{Al}=n_{AlBr_3}=0,4\left(mol\right)\\ n_{Br_2}=\dfrac{3}{2}.0,4=0,6\left(mol\right)\\ a=m_{Al}=0,4.27=10,8\left(g\right)\\ m_{Br_2}=160.0,6=96\left(g\right)\)

2Al+3Br₂-to>2AlBr₃

0,4-----0,6-------0,4 mol

n AlBr3=\(\dfrac{106,8}{267}\)=0,4 mol

=>a=mAl=0,4.27=10,8g

=>mBr₂=0,6.160=96g

Câu 2

Mg+Br₂->MgBr₂

0,08--0,08---0,08 mol

n _MgBr2=\(\dfrac{14,72}{184}\)=0,08 mol

=>a=mMg=0,08.24=1,92g

=>m _Br2=0,08.160=12,8g

Bài 6:

\(n_{Fe\left(OH\right)_3}=\dfrac{21,4}{107}=0,2\left(mol\right)\)

PT: \(Fe\left(OH\right)_3+3HCl\rightarrow FeCl_3+3H_2O\)

_______0,2________0,6______0,2 (mol)

a, \(C\%_{HCl}=\dfrac{0,6.36,5}{200}.100\%=10,95\%\)

b, \(C\%_{FeCl_3}=\dfrac{0,2.162,5}{21,4+200}.100\%\approx14,68\%\)

Bài 7:

\(m_{H_2SO_4}=100.9,8\%=9,8\left(g\right)\Rightarrow n_{H_2SO_4}=\dfrac{9,8}{98}=0,1\left(mol\right)\)

PT: \(ZnO+H_2SO_4\rightarrow ZnSO_4+H_2O\)

______0,1______0,1_______0,1 (mol)

a, \(m_{ZnO}=0,1.81=8,1\left(g\right)\)

b, \(C\%_{ZnSO_4}=\dfrac{0,1.161}{8,1+100}.100\%\approx14,89\%\)