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

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

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

mddH3PO4= 500(g)

=>mH3PO4=500.24,5=122,5(g)

=>nH3PO4=1,25(mol)

PTHH: 4P +5 O2 -to-> 2 P2O5

P2O5 +3 H2O -> 2 H3PO4

nP2O5=1,25/2=0,625(mol)

=>nP=1,25(mol)

=>a=mP=1,25.31=38,75(g)

Chúc em học tốt!

\(n_{H_3PO_4}=\dfrac{500.1.24,5\%}{98}=1,25\left(mol\right)\)

Bảo toàn nguyên tố P : \(n_P=n_{H_3PO_4}=1,25\left(mol\right)\)

=> \(m_P=1,25.31=38,75\left(g\right)\)

4P + 5O2 = 2P2O5 (1)

P2O5 + 3H2O = 2H3PO4 (2)

ta có: m_ddH3PO4= v.D= 500.0,1= 50g

m_H3PO4 = C%.m_dd:100= 24,5%.50:100= 12,25g

n_{H3PO4= \(\dfrac{m}{M}\)} = 12,25:98=0,125mol

thế vào ptr 2 tính được số mol P2O5=0,125.\(\dfrac{1}{2}\)=0,0625mol

thế số mol của P2O5 vào ptr 1 ta tính được số mol P=0,0625.4:2=0,125mol

m_{P= n.M= 0,125.31= 3,875g hay a= 3,875g}

_{nếu sai sửa dùm mình nhé}

\(n_{K_2O}=\dfrac{23.5}{94}=0.25\left(mol\right)\)

\(K_2O+H_2O\rightarrow2KOH\)

\(0.25...................0.5\)

\(C_{M_{KOH}}=\dfrac{0.5}{0.5}=1\left(M\right)\)

\(2KOH+H_2SO_4\rightarrow K_2SO_4+H_2O\)

\(0.5............0.25............0.25\)

\(m_{dd_{H_2SO_4}}=\dfrac{0.25\cdot98}{20\%}=122.5\left(g\right)\)

\(V_{dd_{H_2SO_4}}=\dfrac{122.5}{1.14}=107.5\left(ml\right)=0.1075\left(l\right)\)

\(C_{M_{K_2SO_4}}=\dfrac{0.25}{0.1075+0.5}=0.4\left(M\right)\)

nK2O = 23,5 : 94 = 0,25 (mol)
Vdd = 500ml = 0,5l

PT K2O + H2O ==> 2KOH
TPT 1 1 2 (mol)
TĐB: 0,25 --> 0,5 (mol)

a) C_{M KOH} = 0,5 : 0,5 = 1(M)

b) PT: H2SO4 + 2KOH ==> K2SO4 + 2H2O
TPT: 1 2 1 2 (mol)

a.

\(n_{K_2O}=\dfrac{m}{M}=\dfrac{23,5}{94}=0,25\left(mol\right)\\ PT:\\ K_2O+H_2O\rightarrow2KOH\)

Theo pt: \(n_{KOH}=2n_{K_2O}=0,5\left(mol\right)\)

\(\Rightarrow C_{M\left(KOH\right)}=\dfrac{n}{V}=\dfrac{0,5}{0,5}=1\left(M\right)\)

b.

\(PT:\\ 2KOH+H_2SO_4\rightarrow K_2SO_4+2H_2O\)(2)

Theo pt, \(n_{H_2SO_4}=\dfrac{1}{2}n_{KOH}=0,25\left(mol\right)\)

\(\Rightarrow V_{ddH_2SO_4}=\dfrac{m_{dd}}{D}=\dfrac{n.M:C\%}{D}=\dfrac{0,25.98:20\%}{1,14}=\)107,46(ml)

c.

dd spư có chất tan là K₂SO₄

Theo pt (2), \(n_{K_2SO_4}=\dfrac{1}{2}n_{KOH}=0,25\left(mol\right)\)

=> C_M = \(\dfrac{n}{V}=\dfrac{0,25}{0,107}=2,336\left(M\right)\)

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)

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

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

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

Sửa đề: 9,2 gam Na

\(a,n_{Na_2O}=\dfrac{9,2}{23}=0,4\left(mol\right)\)

PTHH: \(Na_2O+H_2O\rightarrow2NaOH\)

            0,4------------------>0,8

\(\rightarrow C_{M\left(NaOH\right)}=\dfrac{0,8}{0,5}=1,6M\)

\(b,n_{K_2O}=\dfrac{37,6}{94}=0,4\left(mol\right)\)

PTHH: \(K_2O+H_2O\rightarrow2KOH\)

            0,4----------------->0,8

\(\rightarrow C\%_{KOH}=\dfrac{0,8.56}{362,4+37,6}.100\%=11,2\%\)