Bài 1: a) \(n_{Fe_2O_3}=\dfrac{16}{160}=0,1\left(mol\right)\)

\(Fe_2O_3+6HCl\rightarrow2FeCl_3+3H_2O\)

Theo PT: \(n_{HCl}=6n_{Fe_2O_3}=0,6\left(mol\right)\)

=> \(m_{ddHCl}=\dfrac{0,6.36,5}{14,6\%}=150\left(g\right)\)

b) \(n_{FeCl_3}=2n_{Fe_2O_3}=0,2\left(mol\right)\)

=> \(m_{FeCl_3}=0,2.162,5=32,5\left(g\right)\)

a) \(Mg+H_2SO_4\rightarrow MgSO_4+H_2\)

Theo PT: \(n_{H_2}=n_{H_2SO_4}=\dfrac{4,9\%.100}{98}=0,05\left(mol\right)\)

=> \(V_{H_2}=0,05.22,4=1,12\left(l\right)\)

b)Theo PT: \(n_{Mg}=n_{H_2SO_4}=0,05\left(mol\right)\)

=> \(m_{Mg}=0,05.24=1,2\left(g\right)\)

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

a)\(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\)

Theo PT:  \(n_{H_2}=\dfrac{3}{2}n_{Al}=0,3\left(mol\right)\)

=> \(V_{H_2}=0,3.22,4=6,72\left(l\right)\)

b)Theo PT:  \(n_{AlCl_3}=n_{Al}=0,2\left(mol\right)\)

=> \(m_{AlCl_3}=0,2.133,5=26,7\left(g\right)\)

c)Theo PT:  \(n_{HCl}=3n_{Al}=0,6\left(mol\right)\)

=> \(CM_{HCl}=\dfrac{0,6}{0,2}=3M\)

Câu 3 : 

\(m_{ct}=\dfrac{10.80}{100}=8\left(g\right)\)

\(n_{NaOH}=\dfrac{8}{40}=0,2\left(mol\right)\)

a) Hiện tượng : Xuất hiện kết tủa trắng

Pt : \(2NaOH+MgSO_4\rightarrow Na_2SO_4+Mg\left(OH\right)_2|\)

           2                1                  1               1

        0,2                0,1               0,1            0,1

\(n_{Mg\left(OH\right)2}=\dfrac{0,2.1}{2}=0,1\left(mol\right)\)

⇒ \(m_{Mg\left(OH\right)2}=0,1.58=5,8\left(g\right)\)

b) \(n_{MgSO4}=\dfrac{0,2.1}{2}=0,1\left(mol\right)\)

 \(m_{MgSO4}=0,1.120=12\left(g\right)\)

\(m_{ddMgSO}=\dfrac{12.100}{10}=120\left(g\right)\)

c) \(n_{Na2SO4}=\dfrac{0,1.1}{1}=0,1\left(mol\right)\)

⇒ \(m_{Na2SO4}=0,1.142=14,2\left(g\right)\)

\(m_{ddspu}=80+120-5,8=194,2\left(g\right)\)

\(C_{Na2SO4}=\dfrac{14,2.100}{194,2}=7,31\)⁰/₀

 Chúc bạn học tốt

Câu 4 : 

\(n_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right)\)

Pt : \(Fe+H_2SO_4\rightarrow FeSO_4+H_2|\)

         1         1               1           1

        0,2      0,2

\(n_{H2SO4}=\dfrac{0,2.1}{1}=0,2\left(mol\right)\)

\(m_{H2SO4}=0,2.98=19,6\left(g\right)\)

\(m_{ddH2SO4}=\dfrac{19,6.100}{20}=98\left(g\right)\)

\(V_{ddH2SO4}=\dfrac{98}{1,2}\simeq81,67\left(ml\right)\)

 Chúc bạn học tốt

a + 21,7 = 34,7%.(100 + a) → a = 19,908 (gam)

b) Giả sử n_MgSO4.7H2O: b (mol)

Suy ra: 41,608 – 120b = 21,7%.(119,908 – 246b) → b = 0,235

→ m_MgSO4.7H2O = 57,802

\(a,2NaOH+MgSO_4\rightarrow Mg\left(OH\right)_2+Na_2SO_4\\ n_{NaOH}=0,5.1=0,5\left(mol\right)\\ b,n_{Mg\left(OH\right)_2}=\dfrac{0,5}{2}=0,25\left(mol\right)=n_{Na_2SO_4}\\ m_{kt}=m_{Mg\left(OH\right)_2}=58.0,25=14,5\left(g\right)\\ c,V_{ddX}=V_{ddNaOH}+V_{ddMgSO_4}=0,5+0,5=1\left(l\right)\\ C_{MddNa_2SO_4}=\dfrac{0,25}{1}=0,25\left(M\right)\)

lớp 5 ?

$a)$

$Mg+H_2SO_4\to MgSO_4+H_2$

$b)$

$n_{Mg}=\frac{2,4}{24}=0,1(mol)$

Theo PT: $n_{MgSO_4}=n_{Mg}=0,1(mol)$

$\to m_{MgSO_4}=0,1.120=12(g)$

$c)$

$CuO+H_2\xrightarrow{t^o}Cu+H_2O$

Theo PT: $n_{Cu}=n_{H_2}=n_{Mg}=0,1(mol)$

$\to m_{Cu}=0,1.64=6,4(g)$

\(CuSO_4+2KOH\rightarrow Cu\left(OH\right)_2+K_2SO_4\\ MgSO_4+2KOH\rightarrow Mg\left(OH\right)_2+K_2SO_4\\ Đặt:a=n_{CuSO_4}\left(mol\right);b=n_{MgSO_4}\left(mol\right)\left(a,b>0\right)\\ \Rightarrow\left\{{}\begin{matrix}160a+120b=68\\98a+58b=37\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,2\\b=0,3\end{matrix}\right.\\ \%m_{CuSO_4}=\dfrac{0,2.160}{68}.100\%\approx47,059\%\Rightarrow\%m_{MgSO_4}\approx52,941\%\)