10FeSO4 + 2KMnO4 + 8H2SO4 ---> 5Fe2(SO4)3 + K2SO4 + 2MnSO4 + 8H2O

Số mol FeSO4 = 1,39/278 = 0,005 mol.

Theo pt trên số mol KMnO4 = 1/5 số mol FeSO4 = 0,001 mol. Suy ra V = 0,001/0,1 = 0,01 lít = 1  

     Nếu thấy đúng thì k cho mk nhé

Câu 6:

\(m_{dd.bđ}=1,1.200=220\left(g\right)\)

\(n_{FeSO_4.7H_2O}=\dfrac{83,4}{278}=0,3\left(mol\right)\Rightarrow n_{FeSO_4}=0,3\left(mol\right)\)

=> \(C\%_{dd.bđ}=\dfrac{0,3.152}{220}.100\%=20,73\%\)

Câu 7:

\(m_{MgCl_2\left(dd.ở.60^oC\right)}=\dfrac{500.37,5}{100}=187,5\left(g\right)\)

=> \(m_{H_2O}=500-187,5=312,5\left(g\right)\)

Giả sử có a mol MgCl₂.6H₂O tách ra

\(n_{MgCl_2\left(dd.ở.10^oC\right)}=\dfrac{187,5}{95}-a=\dfrac{75}{38}-a\left(mol\right)\)

=> \(m_{MgCl_2\left(dd.ở.10^oC\right)}=95\left(\dfrac{75}{38}-a\right)=187,5-95a\left(g\right)\)

\(n_{H_2O\left(tách.ra\right)}=6a\left(mol\right)\)

\(m_{H_2O\left(dd.ở.10^oC\right)}=312,5-18.6a\)=312,5 - 108a (g)

=> \(S_{10^oC}=\dfrac{187,5-95a}{312,5-108a}.100=53\left(g\right)\)

=> \(a=\dfrac{4375}{7552}\left(mol\right)\)

=> \(m_{MgCl_2.6H_2O}=\dfrac{4375}{7552}.203=117,6\left(g\right)\)

1)

\(m_{ddCuSO_4\left(bd\right)}=1,6.25=40\left(g\right)\)

\(n_{CuSO_4.5H_2O}=\dfrac{11,25}{250}=0,045\left(mol\right)\)

=> \(n_{CuSO_4}=0,045\left(mol\right)\)

\(C_M=\dfrac{0,045}{0,025}=1,8M\)

\(C\%=\dfrac{0,045.160}{40}.100\%=18\%\)

b)

\(m_{CuSO_4}=\dfrac{200.18}{100}=36\left(g\right)\)

\(n_{CuSO_4.5H_2O}=\dfrac{5,634}{250}=0,022536\left(mol\right)\)

n_{CuSO4 (tách ra)} = 0,022536 (mol)

=> \(m_{CuSO_4\left(dd.ở.t^o\right)}=36-0,022536.160=32,39424\left(g\right)\)

\(m_{H_2O\left(bd\right)}=200-36=164\left(g\right)\)

n_{H2O (tách ra)} = 0,022536.5 = 0,11268 (mol)

=> \(m_{H_2O\left(dd.ở.t^o\right)}=164-0,11268.18=161,97176\left(g\right)\)

\(S_{t^oC}=\dfrac{32,39424}{161,97176}.100=20\left(g\right)\)

a) 

\(n_{FeSO_4.7H_2O}=\dfrac{41,7}{278}=0,15\left(mol\right)\)

=> \(n_{FeSO_4}=0,15\left(mol\right)\)

=> \(m_{FeSO_4}=0,15.152=22,8\left(g\right)\)

b) m_{dd sau pha trộn} = 41,7 + 207 = 248,7 (g)

c) \(C\%=\dfrac{22,8}{248,7}.100\%=9,168\%\)

\(V_{dd}=\dfrac{248,7}{1,023}=243,1085\left(ml\right)=0,2431085\left(l\right)\)

\(C_M=\dfrac{0,15}{0,2431085}=0,617M\)

Sửa lại câu c . 

\(n_{H_2SO_4}=\dfrac{49.40}{100}:98=0,2\left(mol\right)\)

\(PTHH:\)

                      \(Fe+H_2SO_4\rightarrow FeSO_4+H_2\uparrow\)

trc p/u :          0,3      0,2     

p/u :                0,2       0,2           0,2         0,2 

sau :                0,1       0             0,2              0,2         

-> Fe dư 

\(m_{ddFeSO_4}=0,3.56+49-0,4=65,4\left(g\right)\) ( ĐLBTKL ) 

\(m_{FeSO_4}=0,2.152=30,4\left(g\right)\)

\(C\%=\dfrac{30,4}{65,4}.100\%\approx46,48\%\)

PTHH :

\(Fe+H_2SO_4\rightarrow FeSO_4+H_2\uparrow\)

0,3      0,3           0,3           0,3 

\(n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)

\(a,m_{Fe}=0,3.56=16,8\left(g\right)\)

\(b,C_M=\dfrac{n}{V}=\dfrac{0,3}{0,2}=1,5M\)

\(c,n_{H_2SO_4}=\dfrac{\dfrac{49.40}{100}}{98}=0,2\left(mol\right)\)

\(\rightarrow n_{FeSO_4}=n_{H_2SO_4}=0,2\left(mol\right)\)

\(m_{FeSO_4}=0,2.152=30,4\left(g\right)\)

\(m_{ddFeSO_4}=49+\left(0,2.56\right)-0,2.2=59,8\left(g\right)\)( định luật bảo toàn khối lượng )

\(C\%=\dfrac{30,4}{59,8}.100\%\approx50,84\%\)

Đáp án C

0,1M