Ta có: \(n_{H_2}=\dfrac{10,08}{22,4}=0,45\left(mol\right)\)

PT: \(Fe+2HCl\rightarrow FeCl_2+H_2\)

a, \(n_{Fe}=n_{H_2}=0,45\left(mol\right)\Rightarrow m_{Fe}=0,45.56=25,2\left(g\right)\)

b, \(n_{HCl}=2n_{H_2}=0,9\left(mol\right)\) \(\Rightarrow C_{M_{HCl}}=\dfrac{0,9}{0,15}=6\left(M\right)\)

c, \(FeCl_2+2NaOH\rightarrow Fe\left(OH\right)_2+2NaCl\)

\(4Fe\left(OH\right)_2+O_2\underrightarrow{t^o}2Fe_2O_3+4H_2O\)

Theo PT: \(n_{Fe_2O_3}=\dfrac{1}{2}n_{Fe\left(OH\right)_2}=\dfrac{1}{2}n_{FeCl_2}=\dfrac{1}{2}n_{H_2}=0,225\left(mol\right)\)

\(\Rightarrow m_{Fe_2O_3}=0,225.160=36\left(g\right)\)

a, Ta có: \(n_{KOH}=0,15.2=0,3\left(mol\right)\)

PT: \(KOH+HCl\rightarrow KCl+H_2O\)

Theo PT: \(n_{HCl\left(pư\right)}=n_{KOH}=0,3\left(mol\right)\)

Mà: HCl dư 15%

\(\Rightarrow n_{HCl}=0,3+0,3.15\%=0,345\left(mol\right)\)

\(\Rightarrow V_{ddHCl}=\dfrac{0,345}{1,5}=0,23\left(l\right)\)

b, Theo PT: \(n_{KCl}=n_{KOH}=0,3\left(mol\right)\)

\(\Rightarrow C_{M_{KCl}}=\dfrac{0,3}{0,15+0,23}\approx0,789\left(M\right)\)

\(C_{M_{HCl\left(dư\right)}}=\dfrac{0,3.15\%}{0,15+0,23}\approx0,118\left(M\right)\)

a ) \(mol_{HCl}=0,5\)

\(\Rightarrow mol_{M\left(OH\right)_2}=0,25\)

Nồng độ mol trong : \(M\left(OH\right)_2=\frac{0,25}{0,5}=1,25M\)

b ) Bảo toàn khối lượng là xong :

Theo thứ tự của PT cân bằng thì : \(m_{M\left(OH\right)_2}+m_{HCl}=m_{MCl_2}+m_{H_2O}\)

\(\Leftrightarrow m_{M\left(OH\right)_2}+18,25=52+9\)

\(\Rightarrow m_{M\left(OH\right)_2}=42,75g\)

\(\Rightarrow m_{M\left(OH\right)_2}=\frac{42,75}{0,25}=171g\)

\(\Rightarrow M\) là \(Bari\left(137\right)\)

c) Nồng độ mol đ sau PƯ sẽ là nồng độ mol của :

\(BaCl_2=\frac{mol_{BaCl_2}}{V_{Ba\left(OH\right)_2}+V_{HCl}}=\frac{0,25}{0,2+0,2}=\frac{0,25}{0,4}=0,625M\)

\(n_{KOH}=0,2.0,2=0,04(mol); n_{H_3PO_4}=0,15.0,15=0,0225(mol)\\ \Rightarrow \dfrac{n_{KOH}}{n_{H_3PO_4}}=\dfrac{0,04}{0,0225} \Rightarrow 1<\dfrac{n_{KOH}}{n_{H_3PO_4}}<2\)

\(\Rightarrow \) Tạo 2 muối \(K_2HPO_4,KH_2PO_4\)

Đặt \(\begin{cases} n_{KH_2PO_4}=x(mol)\\ n_{K_2HPO_4}=y(mol) \end{cases}\)

Bảo toàn K: \(x+2y=0,04(1)\)

Bảo toàn P: \(x+y=0,0225(2)\)

\((1)(2)\Rightarrow \begin{cases} x=0,005(mol)\\ y=0,0175(mol) \end{cases}\\ \Rightarrow \begin{cases} C_{M_{KH_2PO_4}}=\dfrac{0,005}{0,2+0,15}\approx 0,014M\\ C_{M_{K_2HPO_4}}=\dfrac{0,0175}{0,2+0,15}=0,05M \end{cases}\)

Ta có: \(n_{H_2}=\dfrac{10,08}{22,4}=0,45\left(mol\right)\)

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

_____0,3_____0,9______0,2____0,45 (mol)

a, m_Al = 0,3.27 = 8,1 (g)

b, \(C_{M_{HCl}}=\dfrac{0,9}{0,15}=6\left(M\right)\)

\(n_{HCl}=0.2\cdot1.5=0.3\left(mol\right)\)

\(n_{KOH}=0.15\cdot2=0.3\left(mol\right)\)

\(KOH+HCl\rightarrow KCl+H_2O\)

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

\(m_{KCl}=0.3\cdot74.5=22.35\left(g\right)\)

\(C_{M_{KCl}}=\dfrac{0.3}{0.2+0.15}=0.85\left(M\right)\)

\(n_{H_3PO_4}=2.0,1=0,2\left(mol\right)\)

\(n_{NaOH}=2,5.0,15=0,375\left(mol\right)\)

PTHH: NaOH + H₃PO₄ --> NaH₂PO₄ + H₂O

_______0,2<----0,2--------->0,2

NaH₂PO₄ + NaOH --> Na₂HPO₄ + H₂O

0,175<-----0,175------>0,175

=> n_NaH2PO4 = 0,025 (mol)

=> n_Na2HPO4 = 0,175 (mol)

=> \(\left\{{}\begin{matrix}C_{M\left(Na_2HPO_4\right)}=\dfrac{0,175}{0,1+0,15}=0,7M\\C_{M\left(NaH_2PO_4\right)}=\dfrac{0,025}{0,1+0,25}=0,1M\end{matrix}\right.\)

\(n_{KMnO_4} = \dfrac{15,8}{158} = 0,1(mol)\\ n_{HCl} = 0,08.2 = 0,16(mol)\)

2KMnO₄ + 16HCl \(\to\) 2KCl + 2MnCl₂ + 5Cl₂ + 8H₂O

0,02.............0,16...................................0,05..............(mol)

\(n_{NaOH} = 0,2.1,5 = 0,3(mol)\)

2NaOH + Cl₂ \(\to\) NaCl + NaClO + H₂O

0,1..........0,05......0,05......0,05........................(mol)

Vậy :

\(C_{M_{NaCl}} = C_{M_{NaClO}} = \dfrac{0,05}{0,2}= 0,25M\\ C_{M_{NaOH}} = \dfrac{0,3-0,1}{0,2} = 1M\)