\(a.n_{H_2}=\dfrac{3,36}{22,4}=0,15mol\\ 2Al+6HCl\rightarrow2AlCl_3+3H_2\)

0,1        0,3             0,1           0,15

\(m_{Al}=0,1.27=2,7g\\ b.C_{M_{HCl}}=\dfrac{0,3}{0,3}=1M\\ c.C_{\%HCl}=\dfrac{0,3.36,5}{300.1,2}\cdot100=3,04\%\\ d)m_{dd}=2,7+300.1,2-0,15.2=362,4g\\ C_{\%AlCl_3}=\dfrac{0,1.133,5}{362,4}\cdot100=3,68\%\)

\(n_{Ba\left(OH\right)_2}=0,5.0,2=0,1\left(mol\right)\)

PTHH: Ba(OH)₂ + 2HCl -----> BaCl₂ + 2H₂O

Mol:       0,1             0,2

\(m_{HCl}=0,2.36,5=7,3\left(g\right)\Rightarrow m_{ddHCl}=\dfrac{7,3.100\%}{15\%}=\dfrac{146}{3}\left(g\right)\)

\(\Rightarrow V_{ddHCl}=\dfrac{\dfrac{146}{3}}{1,25}=\dfrac{584}{15}=38,9\left(ml\right)\)

\(a,n_{MgCO_3}=\dfrac{8,4}{84}=0,1\left(mol\right)\)

PTHH: \(MgCO_3+2HCl\rightarrow MgCl_2+CO_2\uparrow+H_2O\)

             0,1------->0,2-------->0,1-------->0,1

\(\rightarrow\left\{{}\begin{matrix}V=0,1.22,4=2,24\left(l\right)\\a=\dfrac{0,2}{0,5}=0,4M\end{matrix}\right.\\ b,m_{muối}=0,1.95=9,5\left(g\right)\)

Ta có n_Na2O = \(\dfrac{6,2}{62}\) = 0,1 ( mol )

300 ml = 0,3 lít

=> n_HCl = 0,3 . 1 = 0,3 ( mol )

Na₂O + 2HCl → 2NaCl + H₂O

0,1 0,3

=> Lập tỉ số \(\dfrac{0,1}{1}\):\(\dfrac{0,3}{2}\) = 0,1 < 0,15

=> Sau phản ứng Na₂O hết

HCl còn dư

Ta có m_NaCl = d . V = 1,2 . 300 = 360 ( gam )

=> m_{dung dịch} = m_{tham gia}

= 360 + 6,2

= 366,2 ( gam )

=> m_NaCl = 0,2 . 58,5 = 11,7 ( gam )

=> C%_NaCl = \(\dfrac{11,7}{366,2}\) . 100 \(\approx\) 3,2%

=> m_{HCl dư} = ( 0,3 - 0,2 ) . 36,5 = 3,65 ( gam )

=> C%_{HCl dư} =\(\dfrac{3,65}{366,2}\) . 100 \(\approx\) 1%

a) \(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)

\(n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\)

Theoo PTHH: \(n_{H_2}=n_{Zn}=0,1\left(mol\right)\)

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

b) \(m_{dd\left(sau\right)}=250+6,5-0,1.2=256,3\left(g\right)\)

\(\Rightarrow C\%=\dfrac{0.1.161}{256,3}.100\%\approx0,06\%\)

c) \(V_{dd}=m_{dd}.D=256,3.1,15\approx294\left(ml\right)=0,294\left(l\right)\)

\(\Rightarrow C_M=\dfrac{0,1}{0,294}=0,34M\)

Na2CO3+2HCl--->2NaCl+H2O+CO2

nNa2CO3= 21,2/106= 0,2 mol

Ta có:

=> m HCL=300.1,15 = 345 (g)

Theo pthh

n CO2=n Na2CO3=0,2(mol)

m CO2=0,2.44=8,8(g)

m dd=21,2+345-8,8=357,4(g)

a/ PTHH: 2Al + 6HCl ===> 2AlCl₃ + 3H₂

n_Al = 5,4 / 27 = 0,2 mol

n_HCl = 0,3 x 1 = 0,3 mol

Lập tỉ lệ theo phương trình

=> Al hết, HCl dư

=> nH2 = 0,3 mol

=> V_H2(đktc) = 0,3 x 22,4 = 6,72 lít

b/ Bạn tự làm nhé!! Mình đang có chút việc nên ko trả lời hết đc!! Sorry

có m_ddBa(OH)2 không bn

\(n_K=\dfrac{9,75}{39}=0,25\left(mol\right)\)

\(n_{HCl}=\dfrac{300.7,3\%}{36,5}=0,6\left(mol\right)\)

PTHH: 2K + 2HCl --> 2KCl + H₂

Xét tỉ lệ: \(\dfrac{0,25}{2}< \dfrac{0,6}{2}\) => HCl dư

PTHH: 2K + 2HCl --> 2KCl + H₂

          0,25-->0,25-->0,25-->0,125

=> V_H2 = 0,125.22,4 = 2,8 (l)

m_KCl = 0,25.74,5 = 18,625 (g)

m_{dd sau pư} = 9,75 + 300 - 0,125.2 = 309,5 (g)

m_HCl(dư) = (0,6 - 0,25).36,5 = 12,775 (g)

\(\left\{{}\begin{matrix}C\%_{KCl}=\dfrac{18,625}{309,5}.100\%=6,018\%\\C\%_{HCl}=\dfrac{12,775}{309,5}.100\%=4,128\%\end{matrix}\right.\)