CaCO3 + 2HCl ➝ H2O + CaCl2 + CO2↑(1)

x(mol) ............................................ x(mol)

MgCO3 + 2HCl ➝ MgCl2 + H2O + CO2↑(2)

y(mol).................................................y(mol)

Gọi x,y lần lượt là số mol của CaCO3, MgCO3

Theo đề bài ta có hệ pt \(\left\{{}\begin{matrix}100x+84y=134\\x+y=\frac{3,36}{22,4}=0,15\left(mol\right)\end{matrix}\right.\)

Giải hệ ta được \(\left\{{}\begin{matrix}x=0,05\\y=0,1\end{matrix}\right.\)

\(\Rightarrow m_{CaCO_3}=0,05.100=5\left(g\right)\)

\(\Rightarrow m_{MgCO_3}=0,1.84=8,4\left(g\right)\)

Theo PT (1): \(n_{HCl\left(1\right)}=2n_{CaCO_3}=0,1\left(mol\right)\)

Theo PT (2)\(n_{HCl\left(2\right)}=2n_{MgCO_3}=0,2\left(mol\right)\)

\(\Rightarrow n_{HCl}=0,1+0,2=0,3\left(mol\right)\)

\(\Rightarrow V_{HCl}=\frac{0,3}{1}=0,3\left(l\right)\)

2Al + 6HCl → 2AlCl₃ + 3H₂↑ (1)

Al₂O₃ + 6HCl → 2AlCl₃ + 3H₂O (2)

\(n_{H_2}=\frac{13,44}{22,4}=0,6\left(mol\right)\)

a) Theo pt1: \(n_{Al}=\frac{2}{3}n_{H_2}=\frac{2}{3}\times0,6=0,4\left(mol\right)\)

\(\Rightarrow m_{Al}=0,4\times27=10,8\left(g\right)\)

\(\Rightarrow m_{Al_2O_3}=21-10,8=10,2\left(g\right)\)

b) Theo PT1: \(n_{HCl}=3n_{Al}=3\times0,2=0,6\left(mol\right)\)

\(n_{Al_2O_3}=\frac{10,2}{102}=0,1\left(mol\right)\)

Theo Pt2: \(n_{HCl}=6n_{Al_2O_3}=6\times0,1=0,6\left(mol\right)\)

\(\Rightarrow\Sigma n_{HCl}=0,6+0,6=1,2\left(mol\right)\)

\(\Rightarrow m_{HCl}=1,2\times36,5=43,8\left(g\right)\)

\(\Rightarrow m_{ddHCl}=\frac{43,8}{36\%}=121,67\left(g\right)\)

\(\Rightarrow V_{ddHCl}=\frac{121,67}{1,18}=103,11\left(ml\right)\)

Ý bạn là tìm \(V_{dd_{HCl}}\) đúng không?

\(PTHH:2Al+6HCl\rightarrow2AlCl_3+3H_2\uparrow\\ AlCl_3+3NaOH\rightarrow Al\left(OH\right)_3\downarrow+3NaCl\\ \Rightarrow n_{AlCl_3}=\dfrac{1}{3}n_{NaOH}=0,1\left(mol\right)\\ \Rightarrow n_{HCl}=3n_{AlCl_3}=0,3\left(mol\right)\\ \Rightarrow V_{dd_{HCl}}=\dfrac{0,3}{2}=0,15\left(l\right)\)

a) $Mg + 2HCl \to MgCl_2 + H_2$
b) $n_{H_2} = n_{Mg} = \dfrac{2,4}{24} = 0,1(mol)$
$V_{H_2} = 0,1.22,4 = 2,24(lít)$
c) $n_{HCl} = 2n_{Mg} = 0,2(mol) \Rightarrow V_{dd\ HCl} = \dfrac{0,2}{1} = 0,2(lít)$

\(n_{H_2}=\dfrac{4,958}{22,4}=0,2mol\\ Fe+2HCl\rightarrow FeCl_2+H_2\)

0,2       0,4           0,2           0,2

\(\%m_{Fe}=\dfrac{0,2.56}{20}\cdot100\%=56\%\\ \%m_{Cu}=100\%-56\%=44\%\\ V_{ddHCl}=\dfrac{0,4}{4}=0,1l\)

Cái nèk mà kiu lớp 1 thì tui chết liền

Đề thiếu gì k

1)2Al+6HCl ->2Al2Cl3+3H2

Fe+2HCl->FeCl2+H2

Gọi số mol của Al là x;Fe là y

ta có 2x*23+56y=8.3

3x+y=5.6/22.4

giải ra là xong hết bài 1 r nha

\(MgCO_3+2HCl\rightarrow MgCl_2+H_2O+CO_2\)

\(CO_2+Na\left(OH\right)_2\rightarrow CaCO_3+H_2O\)

b. \(n_{MgCO_3}=\dfrac{21}{84}=0,25mol\) \(\Rightarrow n_{HCl}=2.0,25=0,5mol\)

\(V_{ddHCl}=\dfrac{0,5}{2}=0,25l\)

c. \(n_{CO_2}=n_{MgCO_3}=0,25mol\)

\(n_{CaCO_3}=n_{CO_2}=0,25mol\)

\(\Rightarrow m_{CaCO_3}=0,25.100=25g\)