\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)

\(n_{Al}=\dfrac{3,375}{27}=0,125\)

\(\Rightarrow n_{H_2}=\dfrac{0,125}{2}.3=0,1875mol\)      \(\Rightarrow V_{H_2}=0,1875.22,4=4,2l\)

\(n_{AlCl_3}=n_{Al}=0,125mol\)       \(\Rightarrow m_{AlCl_3}=0,125.133,5=16,6875g\)

 PTHH: \(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\uparrow\)

Làm gộp các phần còn lại

Ta có: \(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\) \(\Rightarrow\left\{{}\begin{matrix}n_{Al_2\left(SO_4\right)_3}=0,1mol\\n_{H_2SO_4}=n_{H_2}=0,3mol\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}V_{H_2}=0,3\cdot22,4=6,72\left(l\right)\\m_{Al_2\left(SO_4\right)_3}=0,1\cdot342=34,2\left(g\right)\\m_{H_2SO_4}=0,3\cdot98=29,4\left(g\right)\end{matrix}\right.\) 

\(n_{Al}=\dfrac{2.7}{27}=0.1\left(mol\right)\)

\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)

\(0.1........0.3............0.1......0.15\)

\(m_{HCl}=0.3\cdot36.5=10.95\left(g\right)\)

\(m_{AlCl_3}=0.1\cdot133.5=13.35\left(g\right)\)

\(V_{H_2}=0.15\cdot22.4=3.36\left(l\right)\)

a) Số mol của Al là

\(n=\dfrac{m}{M}\) = \(\dfrac{4.68}{27}\)= 0.1733 mol

PTHH

            2Al   +   6HCl \(\rightarrow\) 2AlCl₃ + 3H₂

            2 mol                    2 mol       3 mol

         0.1733 mol         0.1733 mol    0.26 mol

 Theo pthh \(n_{AlCl_3}\) = 0,1733 mol

vậy Khối lượng của AlCl₃ là

\(m=M\cdot n\) = 133.5 \(\cdot\) 0,1733 = 23.1355 mol

b) theo pthh \(n_{H_2}\) = 0.26 mol

vậy thể tích của H₂ ở đkc là 

\(V=n\cdot24.79\) = 0.26\(\cdot\)24.79 = 6.4454 (L)

a) \(2Al+6HCl\rightarrow2AlCl_3+3H_2\)

\(n_{Al}=\dfrac{m_{Al}}{M_{Al}}=\dfrac{54}{27}=2\left(mol\right)\)

Theo PTHH: \(n_{H_2}=\dfrac{2.3}{2}=3\left(mol\right)\)

\(\Rightarrow V_{H_2}=n_{H_2}.22,4=3.22,4=67,2l\)

b) \(2H_2+O_2\rightarrow2H_2O\)

\(n_{O_2}=\dfrac{m_{O_2}}{M_{O_2}}=\dfrac{30}{32}=0,94\left(mol\right)\)

Theo PTHH: \(n_{H_2O}=\dfrac{0,94.2}{1}=1,88\left(mol\right)\)

\(\Rightarrow m_{H_2O}=n_{H_2O}.M_{H_2O}=1,88.18=33,84\left(g\right)\)

PTPỨ: Zn + 2HCl ---> ZnCl2 + H2
nZn = 6,5/65 = 0,1 (mol)
Theo ptpứ: nHCl = 2nZn = 0,2 (mol)

=> mHCl = 0,2 x 36,5 = 7,3 (g)
nH2 = nZn = 0,1 (mol)
=> VH2(đktc) = 0,1 x 22,4 = 2,24 (l)
mdd HCl đã dùng = 7,3 x 7,3% = 0,5329(g)

Câu c bạn viết rõ ra đi ak. CM hay C% và của chất j

a) nAl=2,7/27=0,1(mol)

PTHH: 2Al + 6 HCl -> 2 AlCl3 + 3H2

0,1_________0,3___0,1_____0,15(mol)

b) mHCl=0,3.36,5=10,95(g)

c) mAlCl3=0,1.133,5=13,35(g)

d) V(H2,đktc)=0,15.22,4=3,36(l)

a. PTHH: 2Al + 6HCl \(\rightarrow\) 2AlCl\(_3\) + 3H\(_2\)

b. \(n_{Al}=\dfrac{m_{Al}}{M_{Al}}=\dfrac{2,7}{27}=0,1\) (mol)

\(n_{HCl}=\dfrac{0,1.6}{2}=0,3\) (mol)

\(m_{HCl}=n_{HCl}.M_{HCl}=0,3.36,5=10,95\) (g)

c. \(n_{AlCl_3}=\dfrac{0,1.2}{2}=0,1\) (mol)

\(m_{AlCl_3}=n_{AlCl_3}.M_{AlCl_3}=0,1.133,5=13,35\) (g)

d. \(n_{H_2}=\dfrac{0,1.2}{3}=0,15\) (mol)

\(V_{H_2}=n_{H_2}.22,4=0,15.22,4=3,36\) (l)

a) 2Al + 6HCl -> 2AlCl3 + 3H2

b)nAl = 2.7/27=0.1mol

2Al + 6HCl -> 2AlCl3 + 3H2

(mol) 0.1 0.3 0.1 0.15

VH2 = 0.15*22.4=3.36l

mAlCl3= 0.1*133.5=13.35g

=>C%=\(\dfrac{13,35}{500+27}100=2,533\%\)

Đề bài phải là thể tích CO₂ bạn nhé!

a, PT: \(CaCO_3+2HCl\rightarrow CaCl_2+H_2O+CO_2\)

Ta có: \(n_{CaCO_3}=\dfrac{4}{100}=0,04\left(mol\right)\)

\(m_{HCl}=\dfrac{14,6.25}{100}=3,65\left(g\right)\Rightarrow n_{HCl}=\dfrac{3,65}{36,5}=0,1\left(mol\right)\)

Xét tỉ lệ: \(\dfrac{0,04}{1}< \dfrac{0,1}{2}\), ta được HCl dư.

Theo PT: \(n_{CO_2}=n_{CaCO_3}=0,04\left(mol\right)\)

\(\Rightarrow V_{CO_2}=0,04.22,4=0,896\left(l\right)\)

b, Theo PT: \(\left\{{}\begin{matrix}n_{HCl\left(pư\right)}=2n_{CaCO_3}=0,08\left(mol\right)\\n_{CaCl_2}=n_{CaCO_3}=0,04\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow n_{HCl\left(dư\right)}=0,02\left(mol\right)\Rightarrow m_{HCl\left(dư\right)}=0,02.36,5=0,73\left(g\right)\)

\(m_{CaCl_2}=0,04.111=4,44\left(g\right)\)

Bạn tham khảo nhé!

PTHH: \(CaCO_3+2HCl\rightarrow CaCl_2+H_2O+CO_2\uparrow\)

Ta có: \(\left\{{}\begin{matrix}n_{CaCO_3}=\dfrac{4}{100}=0,04\left(mol\right)\\n_{HCl}=\dfrac{14,6\cdot25\%}{36,5}=0,1\left(mol\right)\end{matrix}\right.\)

Xét tỉ lệ: \(\dfrac{0,04}{1}< \dfrac{0,1}{2}\) \(\Rightarrow\) HCl còn dư, CaCO₃ p/ứ hết

\(\Rightarrow\left\{{}\begin{matrix}n_{CO_2}=n_{CaCl_2}=0,04\left(mol\right)\\n_{HCl\left(dư\right)}=0,02\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}V_{CO_2}=0,04\cdot22,4=0,896\left(l\right)\\m_{CaCl_2}=0,04\cdot111=4,44\left(g\right)\\m_{HCl\left(dư\right)}=0,02\cdot36,5=0,73\left(g\right)\end{matrix}\right.\)