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

b, \(n_{Mg}=\dfrac{2,4}{24}=0,1\left(mol\right)\)

Theo PT: \(n_{H_2SO_4}=n_{MgSO_4}=n_{H_2}=n_{Mg}=0,1\left(mol\right)\)

\(\Rightarrow V_{H_2}=0,1.24,79=2,479\left(l\right)\)

c, \(m_{ddH_2SO_4}=\dfrac{0,1.98}{19,6\%}=50\left(g\right)\)

d, Ta có: m _{dd sau pư} = 2,4 + 50 - 0,1.2 = 52,2 (g)

\(\Rightarrow C\%_{MgSO_4}=\dfrac{0,1.120}{52,2}.100\%\approx22,99\%\)

a. PTHH: Zn + H₂SO₄ ---> ZnSO₄ + H₂↑

b. Ta có: \(C_{M_{H_2SO_4}}=\dfrac{n_{H_2SO_4}}{200:1000}=1,5M\)

=> \(n_{H_2SO_4}=0,3\left(mol\right)\)

Ta lại có: \(n_{Zn}=\dfrac{16,25}{65}=0,25\left(mol\right)\)

Ta thấy: \(\dfrac{0,3}{1}>\dfrac{0,25}{1}\)

Vậy H₂SO₄ dư.

Theo PT: \(n_{H_2}=n_{Zn}=0,25\left(mol\right)\)

=> \(V_{H_2}=0,25.22,4=5,6\left(lít\right)\)

c. Theo PT: \(n_{ZnSO_4}=n_{Zn}=0,25\left(mol\right)\)

=> \(m_{ZnSO_4}=0,25.161=40,25\left(g\right)\)

d. Ta có: \(V_{dd_{ZnSO_4}}=0,2\left(lít\right)\)

=> \(C_{M_{ZnSO_4}}=\dfrac{0,25}{0,2}=1,25M\)

\(n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\\ pthh:Zn+2HCl\rightarrow ZnCl_2+H_2\) 
          0,1           0,2            0,1      0,1 
\(m_{HCl}=0,2.36,5=7,3\left(g\right)\\ V_{H_2}=0,1.22,4=2,24l\\ m_{\text{dd}}=6,5+200-\left(0,1.2\right)=206,3g\)  
bài 2 :
\(n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\\ pthh:Mg+2HCl\rightarrow MgCl_2+H_2\) 
          0,2             0,4       0,2              0,2 
\(m_{HCl}=0,4.36,5=14,6g\\ V_{H_2}=0,2.22,4=4,48l\\ m\text{dd}=4,8+200-0,4=204,4g\\ C\%=\dfrac{0,2.136}{204,4}.100\%=13,3\%\)

\(a.Mg+H_2SO_4\rightarrow MgSO_4+H_2\\ b.n_{Mg}=\dfrac{2,4}{24}=0,1mol\\ n_{MgSO_4}=n_{Mg}=0,1mol\\ m_{MgSO_4}=0,1.120=12g\)

c. Nếu thay bằng \(H_2SO_4\) 1M thì lượng Mg sẽ không ta hết.

Nếu Pứ ở \(55^0C\) thì pứ xảy ra nhanh hơn.

a. \(Mg+2HCl\rightarrow MgCl_2+H_2\)

b. \(n_{Mg}=\dfrac{2.4}{24}=0.1mol\)

\(mct_{HCl}=\dfrac{500\times36.5}{100}=182.5g\Rightarrow n_{HCl}=\dfrac{182.5}{36.5}=5mol\)

Ta có: \(\dfrac{0.1}{1}< \dfrac{5}{2}\Rightarrow\) HCl dư

nHCl phản ứng = 0.2 mol => nHCl dư = 5 - 0.2 = 4.8 mol

mHCl dư = \(4.8\times36.5=175.2g\)

c. \(V_{H_2}=0.1\times22.4=2.24l\)

 d. mdd sau phản ứng = \(2.4+500-0.1\times2=502.2g\)

\(C\%_{MgCl_2}=\dfrac{0.1\times95\times100}{502.2}=1.89\%\)

\(n_{Zn}=\dfrac{6.5}{65}=0.1\left(mol\right)\)

\(n_{HCl}=\dfrac{100\cdot14.6\%}{36.5}=0.4\left(mol\right)\)

\(Zn+2HCl\rightarrow ZnCl_2+H_2\)

\(1........2\)

\(0.1......0.4\)

\(LTL:\dfrac{0.1}{1}< \dfrac{0.4}{2}\Rightarrow HCldư\)

\(V_{H_2}=0.1\cdot22.4=2.24\left(l\right)\)

\(m_{\text{dung dịch sau phản ứng}}=6.5+100-0.1\cdot2=106.3\left(g\right)\)

\(C\%ZnCl_2=\dfrac{0.1\cdot136}{106.3}\cdot100\%=12.79\%\)

\(C\%HCl\left(dư\right)=\dfrac{\left(0.4-0.2\right)\cdot36.5}{106.3}\cdot100\%=6.87\%\%\)