a) Fe + 2HCl --> FeCl₂ + H₂

b) \(n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)

PTHH: Fe + 2HCl --> FeCl₂ + H₂

_____0,15<-0,3<----0,15<---0,15

\(\left\{{}\begin{matrix}\%Fe=\dfrac{0,15.56}{12}.100\%=70\%\%\\\%Cu=100\%-70\%=30\%\end{matrix}\right.\)

c) m_HCl = 0,3.36,5 = 10,95 (g)

=> \(m_{dd}=\dfrac{10,95.100}{10}=109,5\left(g\right)\)

d) mdd  = 12 + 109,5 - 0,15.2 = 121,2 (g)

 \(C\%\left(FeCl_2\right)=\dfrac{0,15.127}{121,2}.100\%=15,718\%\)

a: \(Cu+2HCl->CuCl_2+H_2\)

\(Fe+2HCl->FeCl_2+H_2\)

a. PTHH : Mg + 2HCl ➝ MgCl₂ + H₂ (1)

b. theo bài : nH₂ = 3,36 : 22,4 = 0,15 (mol)

theo (1) nMg = nH₂ = 0,15 (mol)

➞ mMg = 0,15 ✖ 24 = 3,6 (g)

➞ %mMg = (3,6 : 5)✖100 = 72%

➞ %mCu = 100% - 72% = 28%

c. theo (1) nHCl = 2nH₂ = 2✖0,15 = 0,3 (mol)

mHCl = 0,3✖36,5 = 10,95(g)

➜mddHCl = (10,95✖100):14,6 = 75(g)

d. dung dịch Y : MgCl₂

mdd_(spư)= 3,6+75-0,3 = 78,3(g)

theo (1) nMgCl₂ = nH₂ = 0,15(mol)

mMgCl₂ = 0,15✖95 = 14,25(g)

C%MgCl₂ = (14,25 : 78,3)✖100 = 18,199%

a) 2Al + 6HCl -> 2AlCl3 + 3H2

Al2O3 + 6HCl -> 2AlCl3 + 3H2O

nH2 = 0,15mol => nAl=0,1mol => mAl=2,7g; mAl2O3 = 10,2g => nAl2O3 = 0,1mol

=>%mAl=20,93% =>%mAl2O3 = 79,07%

b) nHCl = 0,1.3+0,1.6=0,9 mol=>mHCl(dd)=100g

mddY=12,9+100-0,15.2=112,6g

mAlCl3=22,5g=>C%=19,98%

Lượng khí hidro thoát ra là 0,15 mol.

a) Al₂O₃ + 2NaOH \(\rightarrow\) 2NaAlO₂ + H₂O

2Al + 2NaOH + 2H₂O \(\rightarrow\) 2NaAlO₂ + 3H₂.

b) Lượng Al phản ứng bằng 2/3 số mol khí hidro thoát ra và bằng 0,1 mol.

\(\Rightarrow\) %m_nhôm=\(\dfrac{27.0,1}{12,9}\).100%\(\approx\)20,93%, %m_{nhôm oxit}\(\approx\)100%-20,93%\(\approx\)79,07%.

a)

$Fe + 2HCl \to FeCl_2 + H_2$
b)

Theo PTHH : $n_{Fe} = n_{H_2} = \dfrac{4,48}{22,4} = 0,2(mol)$
$\Rightarrow m_{Fe} = 0,2.56 = 11,2(gam)$
$\Rightarrow m_{Cu} = 15,6 - 11,2  = 4,4(gam)$

Chào em, em đăng bài nhầm môn, em đăng sang môn Hóa nha!

Bài 1

\(a)n_{H_2}=\dfrac{3,36}{22,4}=0,15mol\\ Fe+2HCl\rightarrow FeCl_2+H_2\)

0,15      0,3          0,15         0,15

\(\%m_{Fe}=\dfrac{0,15.56}{12}\cdot100\%=70\%\\ \%m_{FeO}=100\%-70\%=30\%\\ b)n_{FeO}=\dfrac{12-0,15.56}{72}=0,05mol\\ FeO+2HCl\rightarrow FeCl_2+H_2O\)

0,05        0,1

\(V_{ddHCl}=\dfrac{0,1+0,3}{2}=0,2l\)

Bài 2

\(a)Na_2O+H_2O\rightarrow2NaOH\\ b)BaO+H_2O\rightarrow Ba\left(OH\right)_2\\ c)BaSO_4?\\ BaO+H_2O\rightarrow Ba\left(OH\right)_2\\ Ba\left(OH\right)_2+CuSO_4\rightarrow BaSO_4+Cu\left(OH\right)_2\\ d)Na_2O+H_2O\rightarrow2NaOH\\ 2NaOH+CuSO_4\rightarrow Cu\left(OH\right)_2+Na_2SO_4\\ e)Na_2O+H_2O\rightarrow2NaOH\\ 2NaOH+FeCl_2\rightarrow Fe\left(OH\right)_2+2NaCl\)