\(a/KOH+HCl\rightarrow KCl+H_2O\\ n_{HCl}=0,25.1,5=0,375mol\\ n_{KOH}=n_{KCl}=n_{HCl}=0,375mol\\ V_{KOH}=\dfrac{0,375}{2}=0,1875l\\ b/C_{M_{KCl}}=\dfrac{0,375}{0,1875+0,25}=\dfrac{6}{7}M\)

a) Fe + 2HCl --> FeCl₂ + H₂

b) \(n_{Fe}=\dfrac{22,4}{56}=0,4\left(mol\right)\)

PTHH: Fe + 2HCl --> FeCl₂ + H₂

_____0,4--->0,8------>0,4--->0,4

=> V_H2 = 0,4.22,4 = 8,96(l)

c) m_HCl = 0,8.36,5 = 29,2 (g)

=> \(m_{dd\left(HCl\right)}=\dfrac{29,2.100}{7,3}=400\left(g\right)\)

mdd (sau pư) = 22,4 + 400 - 0,4.2 = 421,6 (g)

=> \(C\%\left(FeCl_2\right)=\dfrac{127.0,4}{421,6}.100\%=12,05\%\)

\(n_{Al}=\dfrac{10,8}{27}=0,4(mol)\\ a,2Al+3H_2SO_4\to Al_2(SO_4)_3+3H_2\\ \Rightarrow n_{H_2}=n_{H_2SO_4}=0,6(mol);n_{Al_2(SO_4)_3}=0,2(mol)\\ b,V_{H_2}=0,6.22,4=13,44(l)\\ c,m_{dd_{H_2SO_4}}=\dfrac{0,6.98}{9,8\%}=600(g)\\ \Rightarrow C\%_{Al_2(SO_4)_3}=\dfrac{0,2.342}{10,8+600-0,6.2}.100\%=11,22\%\)

a) \(n_{Al}=\dfrac{m}{M}=\dfrac{10,8}{27}=0,4\left(\text{mol}\right)\)

Phương trình hóa học phản ứng

2Al + 3H₂SO₄ ---> Al₂(SO₄)₃ + 3H₂

_{2          :   3                   : 1               : 3} 

0,4        0,6            0,2             0,6

mol       mol           mol             mol

Thể tích khí H₂ sinh ra là 

\(V=n.22,4=0,6.22,4=13,44\left(l\right)\)

\(m_{Al_2\left(SO_4\right)_3}=n.M=0,2.342=68,4\left(g\right)\)

\(m_{H_2SO4}=n.M=0,6.98=58,8\left(g\right)\)

\(m_{H_2}=n.M=0,6.2=1,2\left(g\right)\)

=> \(m_{ddH_2SO_4}=\dfrac{m_{H_2SO_4}.100\%}{C\%}=\dfrac{58,8.100\%}{9,8\%}=600\)(g)

=> \(m_{\text{dd sau pư}}=m_{ddH_2SO_4}+m_{Al}-m_{H_2}\)

= 600 + 10,8 - 1,2 (g) = 609,6 (g) 

=> \(C\%_{Al_2\left(SO_4\right)_3}=\dfrac{m_{Al_2\left(SO_4\right)_3}}{m_{\text{dd sau pư}}}.100\%=\dfrac{68,4}{609,6}.100\%\)=11,22%

\(a,PTHH:KOH+HCl\rightarrow KCl+H_2O\\ b,n_{KOH}=n_{HCl}=2\cdot0,1=0,2\left(mol\right)\\ \Rightarrow V_{dd_{HCl}}=\dfrac{0,2}{2}=0,1\left(l\right)\\ c,C_{M_{KCl}}=\dfrac{0,2}{0,1+0,1}=1M\)

Đổi 100ml = 0,1 lít

Ta có: \(n_{KOH}=2.0,1=0,2\left(mol\right)\)

a. PTHH: \(KOH+HCl--->KCl+H_2O\)

b. Theo PT: \(n_{HCl}=n_{KOH}=0,2\left(mol\right)\)

\(\Rightarrow V_{dd_{HCl}}=\dfrac{0,2}{2}=0,1\left(lít\right)=100\left(ml\right)\)

c. Ta có: \(V_{dd_{KCl}}=V_{dd_{HCl}}=0,1\left(lít\right)\)

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

\(\Rightarrow C_{M_{KCl}}=\dfrac{0,2}{0,1}=2M\)

Bổ sung: \(D_{HCl}=1,18\left(g/ml\right)\)

a) PTHH: \(MgO+2HCl\rightarrow MgCl_2+H_2O\)

b) Ta có: \(\left\{{}\begin{matrix}n_{MgO}=\dfrac{2}{40}=0,05\left(mol\right)\\n_{HCl}=\dfrac{100\cdot1,18\cdot20\%}{36,5}=\dfrac{236}{365}\left(mol\right)\end{matrix}\right.\)

Xét tỉ lệ: \(\dfrac{0,05}{1}< \dfrac{\dfrac{236}{365}}{2}\) \(\Rightarrow\) HCl còn dư, MgO p/ứ hết

\(\Rightarrow n_{MgCl_2}=0,05\left(mol\right)\) \(\Rightarrow C_{M_{MgCl_2}}=\dfrac{0,05}{0,1}=0,5\left(M\right)\)

a) PTHH: \(SO_3+H_2O\rightarrow H_2SO_4\)

Ta có: \(n_{SO_3}=\dfrac{24}{80}=0,3\left(mol\right)=n_{H_2SO_4}\) \(\Rightarrow m_{ddH_2SO_4}=\dfrac{0,3\cdot98}{20\%}=147\left(g\right)\)

\(\Rightarrow V_{ddH_2SO_4}=\dfrac{147}{1,14}\approx128,95\left(ml\right)\)

b) PTHH: \(Fe+H_2SO_4\rightarrow FeSO_4+H_2\uparrow\)

Theo PTHH: \(n_{Fe}=n_{H_2SO_4}=n_{H_2}=0,3\left(mol\right)=n_{FeSO_4}\)

\(\Rightarrow\left\{{}\begin{matrix}m_{Fe}=0,3\cdot56=16,8\left(g\right)\\V_{H_2}=0,3\cdot24,76=7,428\left(l\right)\\m_{FeSO_4}=0,3\cdot152=45,6\left(g\right)\\m_{H_2}=0,3\cdot2=0,6\left(g\right)\end{matrix}\right.\)

Mặt khác: \(m_{dd}=m_{Fe}+m_{ddH_2SO_4}-m_{H_2}=163,2\left(g\right)\)

\(\Rightarrow C\%_{FeSO_4}=\dfrac{45,6}{163,2}\cdot100\%\approx27,94\%\)

\(n_{ZnO}=\dfrac{16.2}{81}=0.2\left(mol\right)\)

\(ZnO+H_2SO_4\rightarrow ZnSO_4+H_2O\)

CM H2SO4 = 0.2/0.1 = 2 (M) 

mZnSO4 = 0.2*161 = 32.2 (g) 

\(n_{K_2O}=\dfrac{1,88}{94}=0,02(mol)\\ a,K_2O+H_2O\to 2KOH\\ b,n_{KOH}=0,04(mol)\\ \Rightarrow C_{M_{KOH}}=\dfrac{0,04}{0,5}=0,08M\\ c,n_{KOH}=0,04.50\%=0,02(mol)\\ KOH+HCl\to KCl+H_2O\\ \Rightarrow n_{HCl}=0,02(mol)\\ \Rightarrow m_{dd_{HCl}}=\dfrac{0,02.36,5}{7,3\%}=10(g)\)

\(n_{HCl}=\dfrac{100.7,3\%}{36,5}=0,2\left(mol\right)\\ a,PTHH:Zn+2HCl\rightarrow ZnCl_2+H_2\\ b,n_{Zn}=n_{H_2}=n_{ZnCl_2}=\dfrac{0,2}{2}=0,1\left(mol\right)\\ m=m_{Zn}=0,1.65=6,5\left(g\right)\\ c,V_{H_2\left(đktc\right)}=0,1.22,4=2,24\left(l\right)\\ d,m_{ddZnCl_2}=6,5+100-0,1.2=106,3\left(g\right)\\ C\%_{ddZnCl_2}=\dfrac{0,1.136}{106,3}.100\approx12,794\%\)