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Ta có: \(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\)
PT: \(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)
Theo PT: \(n_{H_2SO_4}=n_{FeSO_4}=n_{H_2}=n_{Fe}=0,1\left(mol\right)\)
a, \(V_{H_2}=0,1.24,79=2,479\left(l\right)\)
b, \(m_{H_2SO_4}=0,1.98=9,8\left(g\right)\)
\(\Rightarrow m_{ddH_2SO_4}=\dfrac{9,8}{9,8\%}=100\left(g\right)\)
c, Ta có: m dd sau pư = 5,6 + 100 - 0,1.2 = 105,4 (g)
\(\Rightarrow C\%_{FeSO_4}=\dfrac{0,1.152}{105,4}.100\%\approx14,42\%\)
\(n_{Fe2O3}=\dfrac{16}{160}=0,1\left(mol\right)\)
a) Pt : \(Fe_2O_3+3H_2SO_4\rightarrow Fe_2\left(SO_4\right)_3+3H_2O\)
0,1-------->0,3------------>0,1
b) \(m_{ddH2SO4}=\dfrac{0,3.98}{9,8\%}.100\%=300\left(g\right)\)
c) \(m_{ddspu}=16+300=316\left(g\right)\)
\(C\%_{Fe2\left(SO4\right)3}=\dfrac{0,1.400}{316}.100\%=12,66\%\)
Ta có: \(m_{H_2SO_4}=50.9,8\%=4,9\left(g\right)\Rightarrow n_{H_2SO_4}=\dfrac{4,9}{98}=0,05\left(mol\right)\)
PT: \(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)
a, \(n_{Fe}=n_{H_2SO_4}=0,05\left(mol\right)\Rightarrow m_{Fe}=0,05.56=2,8\left(g\right)\)
b, \(n_{FeSO_4}=n_{H_2}=n_{H_2SO_4}=0,05\left(mol\right)\)
Ta có: m dd sau pư = 2,8 + 50 - 0,05.2 = 52,7 (g)
\(\Rightarrow C\%_{FeSO_4}=\dfrac{0,05.152}{52,7}.100\%\approx14,42\%\)
\(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\)
\(PTHH:Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)
(mol)_____0,2____0,2______0,2____0,2__
\(a.V_{H_2}=22,4.0,2=4,48\left(l\right)\)
\(b.m_{ddH_2SO_4}=\dfrac{0,2.98.100}{24,5}=80\left(g\right)\)
\(c.m_{ddspu}=13+80-0,2.2=92,6\left(g\right)\\ \Rightarrow C\%_{ddspu}=\dfrac{0,2.136}{92,6}.100=29,4\left(\%\right)\)
\(n_{Fe_2O_3}=\dfrac{48}{160}=0,3(mol)\\ PTHH:Fe_2O_3+3H_2SO_4\to Fe_2(SO_4)_3+3H_2O\\ \Rightarrow n_{H_2SO_4}=3n_{Fe_2O_3}=0,9(mol)\\ \Rightarrow m_{dd_{H_2SO_4}}=\dfrac{0,9.98}{19,6\%}=450(g)\)
a, \(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)
b, Ta có: \(n_{H_2SO_4}=0,5.1=0,5\left(mol\right)\)
Theo PT: \(n_{Fe}=n_{H_2SO_4}=0,5\left(mol\right)\Rightarrow m_{Fe}=0,5.56=28\left(g\right)\)
c, \(n_{FeSO_4}=n_{H_2SO_4}=0,5\left(mol\right)\Rightarrow C_{M_{FeSO_4}}=\dfrac{0,5}{0,5}=1\left(M\right)\)
\(n_{H_2SO_4}=\dfrac{200.7,35\%}{98}=0,15\left(mol\right)\)
PTHH: 2NaOH + H2SO4 --> Na2SO4 + 2H2O
0,3<----0,15-------->0,15
=> mNaOH = 0,3.40 = 12 (g)
\(m_{dd.NaOH}=\dfrac{12.100}{8}=150\left(g\right)\)
mdd sau pư = 200 + 150 = 350 (g)
mNa2SO4 = 0,15.142 = 21,3 (g)
=> \(C\%_{dd.Na_2SO_4}=\dfrac{21,3}{350}.100\%=6,086\%\)
\(n_{SO_3}=\dfrac{20}{80}=0,25\left(mol\right)\\ PTHH:SO_3+H_2O\rightarrow H_2SO_4\\ Mol:0,25\rightarrow0,25\rightarrow0,25\\ C_{MH_2SO_4}=\dfrac{0,25}{0,5}=0,5M\\ Mg+H_2SO_4\rightarrow MgSO_4+H_2\uparrow\\ Mol:0,25\leftarrow0,25\\ m_{Mg}=0,25.24=6\left(g\right)\)
a) \(n_{CuSO_4}=\dfrac{16}{160}=0,1\left(mol\right)\)
\(C_{M_{ddCuSO_4}}=\dfrac{0,1}{0,2}=0,5M\)
b) \(m_{H_2SO_4}=\dfrac{150.14}{100}=21\left(g\right)\)
Ta có: \(n_{Fe_2O_3}=\dfrac{16}{160}=0,1\left(mol\right)\)
PT: \(Fe_2O_3+3H_2SO_4\rightarrow Fe_2\left(SO_4\right)_3+3H_2O\)
a, \(n_{H_2SO_4}=3n_{Fe_2O_3}=0,3\left(mol\right)\)
\(\Rightarrow m_{H_2SO_4}=0,3.98=29,4\left(g\right)\)
\(\Rightarrow m_{ddH_2SO_4}=\dfrac{29,4}{9,8\%}=300\left(g\right)\)
b, \(n_{Fe_2\left(SO_4\right)_3}=n_{Fe_2O_3}=0,1\left(mol\right)\)
Ta có: m dd sau pư = 16 + 300 = 316 (g)
\(\Rightarrow C\%_{Fe_2\left(SO_4\right)_3}=\dfrac{0,1.400}{316}.100\%\approx12,66\%\)