\(a)Zn+2HCl\rightarrow ZnCl_2+H_2\\ b)n_{Zn}=\dfrac{6,5}{65}=0,1mol\\ Zn+2HCl\rightarrow ZnCl_2+H_2\)
0,1      0,2             0,1             0,1

\(m_{ZnCl_2}=0,1.136=13,6g\\ c)V_{H_2}=0,1.24,79=2,479l\\ d)C_{\%HCl}=\dfrac{0,2.36,5}{200}\cdot100\%=3,65\%\)

\(n_{H_2}=\dfrac{2.479}{24.79}=0.1\left(mol\right)\)

\(Fe+2HCl\rightarrow FeCl_2+H_2\)

\(0.1.......0.2..........0.1......0.1\)

\(m_{Fe}=0.1\cdot56=5.6\left(g\right)\)

\(C_{M_{HCl}}=\dfrac{0.2}{0.2}=1\left(M\right)\)

\(n_{Fe}=\dfrac{7,2}{56}=0,1\left(m\right)\)

PTHH : Fe + 2HCl -> FeCl₂ + H₂

tỉ lệ     :1        2          1           1

số mol:0,1     0,2       0,1        0,1

m_HCl=0,2.36,5=7,3(g)

\(V_{H_2}=0,1.22,4=2,24\left(l\right)\)

\(2Al+3H_2SO_4 \to Al_2(SO_4)_3+3H_2\\ Zn+H_2SO_4 \to ZnSO_4+H_2\\ n_{H_2}=\frac{2,24}{22,4}=0,1(mol)\\ n_{Al}=a(mol)\\ n_{Zn}=b(mol)\\ m_{hh}=27a+65b=3,68(1)\\ n_{H_2}=1,5a+b=0,1(2)\\ (1)(2)\\ a=b=0,04(mol)\\ n_{H_2SO_4}=1,5a+b=1,5.0,04+0,04=0,11mol\\ m_{dd H_2SO_4}=\frac{98.0,1.100\%}{10\%}=98(g)\\ m_{dd}=3,68+98-0,1.2=101,48(g)\)

\(n_{Zn}=\dfrac{9.75}{65}=0.15\left(mol\right)\)

\(Zn+2HCl\rightarrow ZnCl_2+H_2\)

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

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

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

nZn = 6,5/ 65 = 0,1 mol

mHCI = 100. 14,6% = 14,6 (g)

nHCI = 14,6/36,5 = 0,4 (mol)

a ) Theo PTHH nH2 = nZn = 0,1 mol

\(\Rightarrow\)VH2 (điều kiện đạt tiêu chuẩn) = 0,1 . 22,4 = 2,24 (l)

b ) Theo PTHH : nZnCL2 = nZn = 0,1 mol

\(\Rightarrow\)mZnCL2 = 0,1 . 136 = 13,6 (g)

\(\Rightarrow\)m chất sau pứ = mHCI dư + mZnCI2 = 7,3 + 13,6 = 20,9 (g) mH2 = 0,1 . 2 = 0,2 (g)

Áp dụng ĐLBTKL ta có :

mdd ZnCI2 = mZn + mddHCI - mH2 = 6,5 + 100- 0,2 = 106,3 (g)

C%ddZnCI2 = 20,9/ 106,3. 100% = 19,7%

PTHH: \(Zn+2HCl\rightarrow ZnCl_2+H_2\uparrow\)

Ta có: \(n_{Zn}=\dfrac{9,75}{65}=0,15\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}n_{HCl}=0,3\left(mol\right)\\n_{ZnCl_2}=0,15\left(mol\right)=n_{H_2}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}V_{H_2}=0,15\cdot22,4=3,36\left(l\right)\\m_{ZnCl_2}=0,15\cdot136=20,4\left(g\right)\\C_{M_{HCl}}=\dfrac{0,3}{0,2}=1,5\left(M\right)\end{matrix}\right.\)

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

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