\(Al_2O_3+HCl\rightarrow AlCl_3+H_2O\)

\(m_{Al_2O_3}=10,2g\)

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\(m_{AlCl_3}=?;m_{H_2O}?\)

Bài làm

\(n_{AL_2O_3}=\dfrac{m}{M}=\dfrac{10,2}{2.27+3.16}=0,1\left(mol\right)\)

Theo PTHH :

\(n_{AlCl_3}=n_{H_2O}=n_{Al_2O_3}=0,1\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}m_{AlCl_3}=n.M=0,1.\left(27+3.35,5\right)=6,25\left(g\right)\\m_{H_2O}=n.M=0,1.\left(2.1+16\right)=1,8\left(g\right)\end{matrix}\right.\)

Sorry bn ,gửi bài r mk ms kiểm tra lại ,mk chưa cân bằng phương trình nha bn,bh mk cân bằng xg r lm tương tự nha bn

\(AL_2O_3+6HCl\rightarrow2AlCl_3+3H_2O\)

\(n_{Al}=\dfrac{10,2}{102}=0,1\left(mol\right)\)

PTHH: Al₂O₃ + 6HNO₃ ---> 2Al(NO₃)₃ + 3H₂O

           0,1           0,6                0,2             0,3

\(\rightarrow m_{HNO_3}=0,6.63=37,8\left(g\right)\\ m_{ddHNO_3}=\dfrac{37,8}{15\%}=252\left(g\right)\\ m_{dd\left(sau.pư\right)}=252+10,2=262,2\left(g\right)\\ m_{Al\left(NO_3\right)_3}=0,2.213=42,6\left(g\right)\\ C\%_{Al\left(NO_3\right)_3}=\dfrac{42,6}{262,2}=16,25\%\)

n H2SO4=\(\dfrac{10\%.490}{2+32+16.4}=0,5mol\)

n Al2O3 =\(\dfrac{10,2}{27.2+16.3}=0,1mol\)

       \(Al_2O_3+3H_2SO_4->Al_2\left(SO_4\right)_3+3H_2O\)

bđ    0,1............0,5

pư    0,1............0,3..................0,1

spu   0 ................0,2................0,1 

=> sau pư gồm H2SO4 dư , Al2(S04)3 và H2O

m H2SO4 dư = \(0,2.\left(2+32+16.3\right)=19,6g\)

m Al2(SO4)3 = \(0,1\left(27.2+32.3+16.4.3\right)=34,2g\)

m dd = \(490+10,2=500,2g\)

% Al2(SO4)3 = \(\dfrac{34,2}{500,2}.100\sim6,84\%\)

% H2SO4 dư = \(\dfrac{19,6}{500,2}.100\sim3,92\%\)

1.

\(m_{HCl}=\dfrac{10,95.75}{100}=8,2125\left(g\right)\)

\(\Rightarrow n_{HCl}=\dfrac{8,2125}{35,5}=0,225\left(mol\right)\)

\(Fe_2O_3+6HCl\rightarrow2FeCl_3+3H_2O\)

\(\Rightarrow n_{Fe_2O_3}=\dfrac{1}{6}n_{HCl}=0,0375\left(mol\right)\)

\(n_{FeCl_3}=\dfrac{1}{3}n_{HCl}=0,075\left(mol\right)\)

\(\Rightarrow C\%=\dfrac{0,075.162,5}{0,0375.160+75}.100\%=15,05\%\)

2.

\(n_{Al_2O_3}=0,1\left(mol\right);n_{HCl}=0,8\left(mol\right)\)

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

Dễ thấy HCl dư.

\(\Rightarrow n_{AlCl_3}=2n_{Al_2O_3}=0,2\left(mol\right)\)

\(\Rightarrow m_{AlCl_3}=26,7\left(g\right)\)

\(\Rightarrow C\%=\dfrac{26,7}{10,2+200}.100\%=12,7\%\)

a)

$n_{KOH} = \dfrac{44,8}{56} = 0,8(mol)$
$n_{SO_2} = \dfrac{6,72}{22,4} = 0,3(mol)$

Ta thấy : $n_{KOH} : n_{SO_2} = 0,8 : 0,3 = 2,67 > 2$ nên dung dịch sau phản ứng có :  $KOH$ dư ; $K_2SO_3$

b)

$n_{K_2SO_3} = n_{SO_2} = 0,3(mol)$
$\Rightarrow m_{K_2SO_3} = 0,3.158 = 47,4(gam)$

$n_{KOH\ dư} = 0,8 - 0,3.2 = 0,2(mol)$
$\Rightarrow m_{KOH\ dư} = 0,2.56 = 11,2(gam)$

a, Ta có: \(n_{SO_2}=\dfrac{13,44}{22,4}=0,6\left(mol\right)\)

\(n_{KOH}=\dfrac{44,8}{56}=0,8\left(mol\right)\)

\(\Rightarrow\dfrac{n_{KOH}}{n_{SO_2}}=1,3333\)

⇒ Dd sau pư gồm: K₂SO₃ và KHSO₃.

b, PT: \(SO_2+2KOH\rightarrow K_2SO_3+H_2O\)

\(SO_2+KOH\rightarrow KHSO_3\)

Giả sử: \(\left\{{}\begin{matrix}n_{K_2SO_3}=x\left(mol\right)\\n_{KHSO_3}=y\left(mol\right)\end{matrix}\right.\)

Theo PT: \(\left\{{}\begin{matrix}x+y=0,6\\2x+y=0,8\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=0,2\left(mol\right)\\y=0,4\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}m_{K_2SO_3}=0,2.158=31,6\left(g\right)\\m_{KHSO_3}=0,4.120=48\left(g\right)\end{matrix}\right.\)

Bạn tham khảo nhé!

Ta có: \(n_{H_2}=\dfrac{1,12}{22,4}=0,05\left(mol\right)\)

PTHH:

Mg + 2HCl ---> MgCl₂ + H₂ (1)

MgO + 2HCl ---> MgCl₂ + H₂O (2)

Theo PT₍₁₎: \(n_{Mg}=n_{H_2}=0,05\left(mol\right)\)

=> \(m_{Mg}=0,05.24=1,2\left(g\right)\)

=> \(\%_{m_{Mg}}=\dfrac{1,2}{9,2}.100\%=13,04\%\)

\(\%_{m_{MgO}}=100\%-13,04\%=86,96\%\)

Các chất gồm: NaHCO₃, AlCl₃, NaHSO₄, Al

ĐÁP ÁN C

Đáp án C

Các chất gồm: NaHCO₃, AlCl₃, NaHSO₄, Al