Gọi số mol FeSO₄.7H₂O là a (mol)

m_dd = 278 + 278a (g)

\(n_{FeSO_4}=a\left(mol\right)\) 

=> \(m_{FeSO_4}=152a\left(g\right)\)

=> \(C\%=\dfrac{152a}{278a+278}.100\%=4\%\)

=> a = \(\dfrac{139}{1761}\left(mol\right)\)

=> \(m_{FeSO_4.7H_2O}=\dfrac{139}{1761}.278=21,943\left(g\right)\)

gọi khối lượng muối FeSO4.7H2O là x(g)

=> mFeSO4.7H2O = 152x/270 = ( x + 270 ) .4 / 100 = 20,65 g

Gọi khối lượng muối là a (gam)

Ta có: \(\dfrac{a}{a+270}=4\%\) \(\Rightarrow a=m_{FeSO_4.7H_2O}=11,25\left(g\right)\)

Bài 4 nha bạn

1)

mFeSO4.7H2O = x gam

—> mFeSO4 = 152x/278 = (x + 278).4%

—> x = 21,94 gam

m FeSO4=\(\dfrac{27,8.152}{278}=15,2g\)

=>m dd=\(\dfrac{15,2}{\left(9\%\right)}=\dfrac{1520}{9}\)

m H2O=\(\dfrac{1520}{9}\)-27,8=141g

->Vh2O=141ml

b)m muối = x (g)

m FeSO4=\(15,2+\dfrac{152x}{278}=20\%\left(x+\dfrac{1520}{9}\right)\)

=>x=53,57g

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

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

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

a) \(m_{HCl}=200\cdot7,3\%=14,6\left(g\right)\)

b) \(n_{NaOH}=0,5\cdot1=0,5\left(mol\right)\) \(\Rightarrow m_{NaOH}=0,5\cdot40=20\left(g\right)\)

c) \(n_{CuSO_4}=0,2\cdot1,5=0,3\left(mol\right)\) \(\Rightarrow m_{CuSO_4}=0,3\cdot160=48\left(g\right)\)

d) Bạn xem lại đề !

a) mHCl=200⋅7,3%=14,6(g)mHCl=200⋅7,3%=14,6(g)

b) nNaOH=0,5⋅1=0,5(mol)nNaOH=0,5⋅1=0,5(mol) ⇒mNaOH=0,5⋅40=20(g)⇒mNaOH=0,5⋅40=20(g)

c) nCuSO4=0,2⋅1,5=0,3(mol)nCuSO4=0,2⋅1,5=0,3(mol) ⇒mCuSO4=0,3⋅160=48(g)⇒mCuSO4=0,3⋅160=48(g)

d) Bạn xem lại đề !