\(n_{Fe_2O_3}=\dfrac{12}{160}=0,075\left(mol\right)\\a, Fe_2O_3+3H_2\rightarrow\left(t^o\right)2Fe+3H_2O\\ b,n_{H_2}=3.0,075=0,225\left(mol\right)\\ V_{H_2\left(đkc\right)}=24,79.0,225=5,57775\left(l\right)\\ c,n_{Fe}=2.0,075=0,15\left(mol\right)\\ m_{Fe}=0,15.56=8,4\left(g\right)\)

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

b, \(n_{H_2}=\dfrac{7,437}{24,79}=0,3\left(mol\right)\)

Theo PT: \(n_{Al}=\dfrac{2}{3}n_{H_2}=0,2\left(mol\right)\Rightarrow m_{Al}=0,2.27=5,4\left(g\right)\)

c, \(n_{HCl}=2n_{H_2}=0,6\left(mol\right)\)

\(\Rightarrow V_{ddHCl}=\dfrac{0,6}{2}=0,3\left(l\right)=300\left(ml\right)\)

\(n_{H_2}=\dfrac{7,437}{24,79}=0,3\left(mol\right)\\a, 2Al+6HCl\rightarrow2AlCl_3+3H_2\\ n_{Al}=n_{AlCl_3}=\dfrac{2}{3}.0,3=0,2\left(mol\right)\\ b,m=m_{Al}=0,2.27=5,4\left(g\right)\\ c,n_{HCl}=\dfrac{6}{3}.0,3=0,6\left(mol\right)\\ V=V_{ddHCl}=\dfrac{0,6}{2}=0,3\left(l\right)=300\left(ml\right)\)

\(n_{NaOH}=0,5.2=1\left(mol\right)\)

PT: \(NaOH+HNO_3\rightarrow NaNO_3+H_2O\)

Theo PT: \(n_{HNO_3}=n_{NaNO_3}=n_{NaOH}=1\left(mol\right)\)

a, \(C_{M_{HNO_3}}=\dfrac{1}{0,3}=\dfrac{10}{3}\left(M\right)\)

b, \(C_{M_{NaNO_3}}=\dfrac{1}{0,5+0,3}=1,25\left(M\right)\)

\(n_{NaOH}=0,5.2=1\left(mol\right)\\ PTHH:NaOH+HNO_3\rightarrow NaNO_3+H_2O\\ a,n_{HNO_3}=n_{NaOH}=1\left(mol\right)\\ C_{MddHNO_3}=\dfrac{1}{0,3}=\dfrac{10}{3}\left(M\right)\\ b,V_{ddsau}=0,5+0,3=0,8\left(l\right)\\ n_{NaNO_3}=n_{NaOH}=1\left(mol\right)\\ C_{MddNaNO_3}=\dfrac{1}{0,8}=1,25\left(M\right)\)

\(n_{Fe_3O_4}=\dfrac{46,4}{232}=0,2\left(mol\right)\\PTHH:Fe_3O_4+4H_2SO_4\rightarrow FeSO_4+Fe_2\left(SO_4\right)_3+4H_2O\\ n_{H_2SO_4}=4.0,2=0,8\left(mol\right)\\ n_{FeSO_4}=n_{Fe_2\left(SO_4\right)_3}=n_{Fe_3O_4}=0,2\left(mol\right)\\ a,m_{Fe_2\left(SO_4\right)_3}=0,2.400=80\left(g\right)\\ m_{FeSO_4}=152.0,2=30,4\left(g\right)\\ b,C\%_{ddH_2SO_4}=\dfrac{0,8.98}{500}.100\%=15,68\%\)

\(n_{Fe_3O_4}=\dfrac{46,4}{232}=0,2\left(mol\right)\)

PT: \(Fe_3O_4+4H_2SO_4\rightarrow FeSO_4+Fe_2\left(SO_4\right)_3+4H_2O\)

a, Theo PT: \(n_{FeSO_4}=n_{Fe_2\left(SO_4\right)_3}=n_{Fe_3O_4}=0,2\left(mol\right)\)

⇒ m_FeSO4 = 0,2.152 = 30,4 (g)

m_Fe2(SO4)3 = 0,2.400 = 80 (g)

Theo PT: \(n_{H_2O}=n_{H_2SO_4}=4n_{Fe_3O_4}=0,8\left(mol\right)\)

\(\Rightarrow m_{H_2O}=0,8.18=14,4\left(g\right)\)

b, \(C\%_{H_2SO_4}=\dfrac{0,8.98}{500}.100\%=15,68\%\)

\(n_{ZnCO_3}=\dfrac{75}{125}=0,6\left(mol\right);n_{CO_2\left(TT\right)}=\dfrac{12,395}{24,79}=0,5\left(mol\right)\\ ZnCO_3\rightarrow\left(t^o\right)ZnO+CO_2\\ n_{CO_2\left(LT\right)}=n_{ZnCO_3}=0,6\left(mol\right)\\ H=\dfrac{0,5}{0,6}.100\%\approx83,333\%\)

em cảm ơn ạ

\(n_{Zn}=\dfrac{65}{65}=1mol\\ Zn+H_2SO_4\rightarrow ZnSO_4+H_2\\ n_{H_2}=n_{Zn}=1mol\\ V_{H_2,đktc}=1.22,4=22,4l\)

\(1.M_{MgCl_2}=24+35,5\cdot2=95g/mol\\ 2.M_{BaCO_3}=137+12+16\cdot3=197g/mol\\ 3.M_{Cr_2O_7}=52\cdot2+16\cdot7=216g/mol\\ 4.M_{KMnO_4}=39+55+16\cdot4=158g/mol\\ 5.M_{Fe_2\left(SO_4\right)_3}=56\cdot2+\left(32+16\cdot4\right)\cdot3=400g/mol\\ 6M_{Al\left(OH\right)_3}=27+\left(16+1\right)\cdot3=78g/mol\\ 7.M_{NaBr}=23+80=103g/mol\\ 8.M_{ZnS}=65+32=97g/mol\\ 9.M_{Hg\left(NO_3\right)_2}=201+\left(14+16\cdot3\right)\cdot2=325g/mol\\ 10.M_{PbCO_3}=207+12+16\cdot3=267g/mol\)

a, \(MgO+H_2SO_4\rightarrow MgSO_4+H_2O\)

b, Ta có: \(m_{H_2SO_4}=200.9,8\%=19,6\left(g\right)\)

\(\Rightarrow n_{H_2SO_4}=\dfrac{19,6}{98}=0,2\left(mol\right)\)

Theo PT: \(n_{MgO}=n_{MgSO_4}=n_{H_2SO_4}=0,2\left(mol\right)\)

\(\Rightarrow m_{MgO}=0,2.40=8\left(g\right)\)

c, Ta có: m _{dd sau pư} = 8 + 200 = 208 (g)

\(\Rightarrow C\%_{MgSO_4}=\dfrac{0,2.120}{208}.100\%\approx11,54\%\)

\(a,3Fe+2O_2\rightarrow\left(t^o\right)Fe_3O_4\\ n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\\ b,n_{Fe_3O_4}=\dfrac{1}{3}.0,1=\dfrac{1}{30}\left(mol\right)\\ a=m_{Fe_3O_4}=\dfrac{232}{30}=\dfrac{116}{15}\left(g\right)\\ c,n_{O_2}=\dfrac{2}{3}.0,1=\dfrac{1}{15}\left(mol\right)\\ V=V_{O_2\left(đkc\right)}=\dfrac{1}{15}.24,79=1,65266667\left(l\right)\)