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a, \(n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
\(n_{Fe}=n_{H_2}=0,15\left(mol\right)\Rightarrow m_{Fe}=0,15.56=8,4\left(g\right)\)
b, \(n_{Na_2O}=\dfrac{6,2}{62}=0,1\left(mol\right)\)
\(Na_2O+H_2O\rightarrow2NaOH\)
\(n_{NaOH}=2n_{Na_2O}=0,2\left(mol\right)\Rightarrow C_{M_{NaOH}}=\dfrac{0,2}{0,5}=0,4\left(M\right)\)
\(a,n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)
PTHH:
\(Fe+2HCl\rightarrow FeCl_2+H_2\uparrow\)
0,15 0,3 0,15 0,15
\(m_{Fe}=0,15.56=8,4\left(g\right)\)
\(a,n_{Na_2O}=\dfrac{6,2}{62}=0,1\left(mol\right)\)
PTHH :
\(Na_2O+H_2O\rightarrow2NaOH\)
0,1 0,1 0,2
\(C_{M\left(A\right)}=\dfrac{0,2}{0,5}=0,4\left(M\right)\)
Ta có: \(n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\)
PT: \(Mg+2HCl\rightarrow MgCl_2+H_2\)
a, \(n_{H_2}=n_{Mg}=0,2\left(mol\right)\)
\(\Rightarrow V_{H_2}=0,2.22,4=4,48\left(l\right)\)
b, \(n_{HCl}=2n_{H_2}=0,4\left(mol\right)\)
\(\Rightarrow C_{M_{HCl}}=\dfrac{0,4}{0,2}=2\left(M\right)\)
c, PT: \(HCl+KOH\rightarrow KCl+H_2O\)
Theo PT: \(n_{KOH}=n_{HCl}=0,4\left(mol\right)\)
\(\Rightarrow m_{ddKOH}=\dfrac{0,4.56}{5,6\%}=400\left(g\right)\)
\(\Rightarrow V_{ddKOH}=\dfrac{400}{1,045}\approx382,78\left(ml\right)\)
\(n_{H_2}=\dfrac{2,24}{22,4}=0,1(mol)\\ a,Fe+2HCl\to FeCl_2+H_2\\ \Rightarrow n_{Fe}=0,1(mol);n_{HCl}=0,2(mol)\\ \Rightarrow m_{Fe}=0,1.56=5,6(g)\\ C_{M_{HCl}}=\dfrac{0,2}{0,25}=0,8M\\ b,m_{dd_{HCl}}=250.1,12=280(g)\\ n_{FeCl_2}=0,1(mol)\\ \Rightarrow C\%_{FeCl_2}=\dfrac{0,1.127}{5,6+280-0,1.2}.100\%=4,45\%\)
Ta có: \(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\)
a. PTHH: Cu + HCl ---x--->
Fe + 2HCl ---> FeCl2 + H2
b. Ta có Cu không phản ứng nên chất rắn sau phản ứng là Cu
=> A = mCu = 6,4(g)
Theo PT: \(n_{H_2}=n_{Fe}=0,1\left(mol\right)\)
=> \(V=V_{H_2}=0,1.22,4=2,24\left(lít\right)\)
c. Hình như thiếu đề nhé.
Ta có: \(n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\)
a. PTHH: Zn + H2SO4 ---> ZnSO4 + H2
b. Theo PT: \(n_{H_2}=n_{Zn}=0,1\left(mol\right)\)
=> \(V_{H_2}=0,1.22,4=2,24\left(lít\right)\)
c. Theo PT: \(n_{H_2SO_4}=n_{Zn}=0,1\left(mol\right)\)
=> \(m_{H_2SO_4}=0,1.98=9,8\left(g\right)\)
Ta có: \(C_{M_{H_2SO_4}}=\dfrac{9,8}{m_{dd_{H_2SO_4}}}.100\%=25\%\)
=> \(m_{dd_{H_2SO_4}}=39,2\left(g\right)\)
Ta có: \(m_{H_2}=0,1.2=0,2\left(g\right)\)
=> \(m_{dd_{ZnSO_4}}=6,5+39,2-0,2=45,5\left(g\right)\)
Theo PT: \(n_{ZnSO_4}=n_{Zn}=0,1\left(mol\right)\)
=> \(m_{ZnSO_4}=0,1.161=16,1\left(g\right)\)
=> \(C_{\%_{ZnSO_4}}=\dfrac{16,1}{45,5}.100\%=35,4\%\)
\(n_{BaSO_4}=\dfrac{58.25}{233}=0.25\left(mol\right)\)
\(BaCl_2+SO_3+H_2O\rightarrow BaSO_4+2HCl\)
\(0.25........0.25.......................0.25........0.5\)
\(V_{SO_3}=0.25\cdot22.4=5.6\left(l\right)\)
\(m_{dd_{BaCl_2}}=\dfrac{0.25\cdot208}{20\%}=260\left(g\right)\)
\(m_{dd}=m_{SO_3}+m_{dd_{BaCl_2}}-m_{BaSO_4}=0.25\cdot80+260-58.25=221.75\left(g\right)\)
\(C\%_{HCl}=\dfrac{0.5\cdot36.5}{221.75}\cdot100\%=8.2\%\)
