a, \(n_{Fe}=\dfrac{8,4}{56}=0,15\left(mol\right)\)

\(n_{HCl}=0,2.2=0,4\left(mol\right)\)

PT: \(Fe+2HCl\rightarrow FeCl_2+H_2\)

Xét tỉ lệ: \(\dfrac{0,15}{1}< \dfrac{0,4}{2}\), ta được HCl dư.

Theo PT: \(n_{H_2}=n_{Fe}=0,15\left(mol\right)\Rightarrow V_{H_2}=0,15.22,4=3,36\left(l\right)=3360\left(ml\right)\)

b, Theo PT: \(\left\{{}\begin{matrix}n_{FeCl_2}=n_{Fe}=0,15\left(mol\right)\\n_{HCl\left(pư\right)}=2n_{Fe}=0,3\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow n_{HCl\left(dư\right)}=0,4-0,3=0,1\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}C_{M_{FeCl_2}}=\dfrac{0,15}{0,2}=0,75\left(M\right)\\C_{M_{HCl\left(dư\right)}}=\dfrac{0,1}{0,2}=0,5\left(M\right)\end{matrix}\right.\)

c, Ta có: \(m_{ddHCl}=1,25.200=250\left(g\right)\)

⇒ m _{dd sau pư} = 8,4 + 250 - 0,15.2 = 258,1 (g)

\(\Rightarrow\left\{{}\begin{matrix}C\%_{FeCl_2}=\dfrac{0,15.127}{258,1}.100\%\approx7,38\%\\C\%_{HCl\left(dư\right)}=\dfrac{0,1.36,5}{258.1}.100\%\approx1,41\%\end{matrix}\right.\)

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

a.250ml=0,25l ; nHCl=0,25.1,5=0,375mol

KOH+HCl->KCl+H2O

1mol 1mol 1mol

0,375 0,375 0,375

VKOh=0,375/2=0,1875l

b.CM KCL=0,375/0,25=1,5M

c.NaOH+HCL=NaCl+H2O

1mol 1mol

0,375 0,375

mdd NaOH=0,375.40.100/10=150g

n_HCl = C_M . V = 1,5 . 0,25 = 0,375 mol

PTHH: KOH + HCl -------> KCl + H₂O

Pt: 1 1 1 1 (mol)

Pư 0,375 <-0,375 -----> 0,375-> 0,375 (mol)

a) V_KOH = \(\dfrac{n}{C_M}\)= 0,1875 l

b) C_M KCl = \(\dfrac{n}{V}\)= \(\dfrac{0,375}{0,25+0,1875}\)\(\approx\)0,86 M

c) PTHH: NaOH + HCL ----> NaCl + H₂O

Pt: 1 1 1 1 (mol)

Pư 0,375 <- 0,375 ------>0,375 --> 0,375 (mol)

m_ctNaOH = n . M = 15 g

m_{ddNaOH = \(\dfrac{15.100\%}{10\%}\)=} 150 g

:3

a)

Coi V dd HCl = 100(ml)

m dd HCl = 1,25.100 = 125(gam)

n HCl = 125.7,3%/36,5 = 0,25(mol)

[H⁺ ] = [Cl^- ] = CM HCl = 0,25/0,1 = 2,5M

b)

n Al = 0,235(mol)

2Al + 6HCl $\to$ 2AlCl3 + 3H2

n HCl pư = 3n Al = 0,705(mol)

n HCl dư = 0,4.2 - 0,705 = 0,095(mol)

[H⁺ ] = CM HCl dư = 0,095/0,4 = 0,2375M

pH = -log([H⁺ ]) = 0,624

$n_{CuO} = \dfrac{8}{80} = 0,1(mol) ; n_{HCl} = 0,15.2 = 0,3(mol)$
$CuO + 2HCl \to CuCl_2 + H_2O$
Ta thấy : 

$n_{CuO} : 1 < n_{HCl} : 2$ nên HCl dư

$n_{CuCl_2} = n_{CuO} = 0,1(mol)$
$n_{HCl\ pư} = 2n_{CuO} = 0,2(mol) \Rightarrow n_{HCl\ dư} = 0,3 - 0,2 = 0,1(mol)$
$C_{M_{CuCl_2}} = \dfrac{0,1}{0,15} = 0,67M$
$C_{M_{HCl}} = \dfrac{0,1}{0,15} = 0,67M$

a) \(n_{CH_3COOH}=0,1.0,3=0,03\left(mol\right)\)

PTHH: CH₃COOH + NaOH --> CH₃COONa + H₂O

                  0,03---->0,03--------->0,03

=> \(V_{dd.NaOH}=\dfrac{0,03}{1,5}=0,02\left(l\right)\)

b) m_CH3COONa = 0,03.82 = 2,46 (g)

c) \(C_{M\left(CH_3COONa\right)}=\dfrac{0,03}{0,1+0,02}=0,25M\)

Anh bổ sung câu c)

\(C_{MddNa_2SO_4}=\dfrac{0,25}{0,09879+0,5}=0,4175\left(M\right)\)