nNaOH=0,3.1+0,2.1,5=0,6(mol)

CM= 0,6/(0,3+0,2)=1,2M

m=1,05.(300+200)=525(g)

mNaOH=0,6.40=24(g)

C%=24/525.100%=4,57%

* Ta có: \(\left\{{}\begin{matrix}n_{NaOH}=0,4\left(mol\right)\\n_{KOH}=0,2\left(mol\right)\end{matrix}\right.\)\(\Rightarrow n_{OH^-}=0,6\left(mol\right)\)

\(H^+\left(0,6\right)+OH^-\left(0,6\right)\rightarrow H_2O\)

Ta có: \(n_{H^+}=2.n_{H_2SO_4}=0,6\left(mol\right)\)

\(\Rightarrow n_{H_2SO_4}=0,3\left(mol\right)\)

\(\Rightarrow C_{M_{H_2SO_4}}=\dfrac{0,3}{0,3}=1\left(M\right)\)

* Ta có: \(\left\{{}\begin{matrix}n_{H_2SO_4}=0,4\left(mol\right)\\n_{HCl}=0,05\left(mol\right)\end{matrix}\right.\)\(\Rightarrow n_{H^+}=2n_{H_2SO_4}+n_{HCl}=0,4.2+0,05=0,85\left(mol\right)\)

\(OH^-\left(0,85\right)+H^+\left(0,85\right)\rightarrow H_2O\)

Ta có: \(n_{OH^-}=n_{NaOH}=0,85\left(mol\right)\)

\(\Rightarrow C_{M_{NaOH}}=\dfrac{0,85}{0,3}=\dfrac{17}{6}\left(M\right)\)

\(n_{NaOH}=0,2.1=0,2\left(mol\right)\\ n_{H_2SO_4}=0,3.1,5=0,45\left(mol\right)\)

\(2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O\)

0,2------->0,1--------->0,1

Xét \(\dfrac{0,2}{2}< \dfrac{0,45}{1}\Rightarrow\) \(H_2SO_4\)dư

Trong dung dịch D có:

\(\left\{{}\begin{matrix}n_{H_2SO_4}=0,45-0,1=0,35\left(mol\right)\\n_{Na_2SO_4}=0,1\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}CM_{H_2SO_4}=\dfrac{0,35}{0,5}=0,7M\\CM_{Na_2SO_4}=\dfrac{0,1}{0,5}=0,2M\end{matrix}\right.\)

b

\(Ca\left(OH\right)_2+H_2SO_4\rightarrow CaSO_4+2H_2O\)

0,35<---------0,35

\(V_{Ca\left(OH\right)_2}=\dfrac{0,35.74}{1,2}=\dfrac{259}{12}\approx21,58\left(ml\right)\\ \Rightarrow V_{dd.Ca\left(OH\right)_2}=\dfrac{\dfrac{259}{12}.100\%}{10\%}=\dfrac{1295}{6}\approx215,83\left(ml\right)\)

\(V_{ddNaOH\left(tổng\right)}=400+200=600\left(ml\right)=0,6\left(l\right)\\ n_{NaOH\left(tổng\right)}=0,4.0,5+0,2.1,5=0,5\left(mol\right)\\ C_{MddNaOH\left(cuối\right)}=\dfrac{0,5}{0,6}\approx0,833\left(M\right)\)

n_{NaOH trong dd NaOH 1M}=0,3(mol)

n_{NaOH trong dd NaOH 1,5M}=0,3(mol)

C_M=\(\dfrac{0,3+0,3}{0,5}=1,2M\)

C%=\(\dfrac{40.1,2}{10.0,5}=9,6\%\)

Tính C% của dung dịch thu được:

Ta có: m_{d d NaOH(1)}=V.D=500.1,2=600(g)

V_{d d NaOH(1)}=500ml=0,5 (lít)

=> n_NaOH(1)=C_M.V=2.0,5=1 (mol)

=> m_NaOH(1)=n_NaOH.M=1.40=40(gam)

Ta có: m_{d d NaOH(2)}=V.D=300.1,1=330(g)

V_{d d NaOH(2)}=300ml=0,3 (lít)

=> n_NaOH(2)=C_M.V=0,5.0,3=0,15(mol)

=> m_NaOH(2)=n.M=0,15.40=6(gam)

=> m_{NaOH mới}=m_NaOH(1) + m_NaOH(2)=40+6=46(gam)

m_{d d NaOH mới}=m_{d d NaOH(1)} + m_{d d NaOH(2)}=600+330=930(gam)

=> \(C\%_{ddsauphanung}=\dfrac{m_{NaOHmới}.100\%}{m_{ddNaOHmoi}}=\dfrac{46.100}{930}\approx4,95\left(\%\right)\)

Tính C_M của dung dịch thu được :

Ta có: V_{d d NaOH mới}=V_{d d NaOH(1)} + V_{d d NaOH(2)}=0,5+0,3=0,8(lít)

n_{d d NaOH mới}= n _{d d NaOH(1)} + n _{d d NaOH(2)}= 1 + 0,15=1,15(mol)

=> \(C_M=\dfrac{n}{V}=\dfrac{1,15}{0,8}\approx1,44\left(M\right)\)

mNaOH trong dd 2 =200*1,33*30/100=79,8g

suy ra nNaOH= 0,5*5+79,8/40=4,495mol

Vdd=0,5+0,2=0,7(l)

suy ra Cmdd=4,495/0,7=899/140M

@Azue

mình sửa thế này đúng ko

Bài 1 : 15,3g hh ( FeO , ZnO ) + CO dư => thu được rắn A . Biết rằng 80% rắn A này = 9,68g

\(^nNaOH=1.0,3=0,3\left(mol\right)\)

     \(NaOH+HCl\rightarrow NaCl+H_2O\)

mol 0,3           0,3         0,3

a) \(CM_{HCl}=\dfrac{0,3}{0,2}=1,5M\)

b) \(CM_{d^2saupứ}=\dfrac{0,3}{0,3+0,2}=0,6M\)

Chúc bạn học tốt!!!

a) \(n_{NaOH}=0,3.1=0,3\left(mol\right)\)

\(NaOH+HCl\rightarrow NaCl+H_2O\)

\(n_{HCl}=n_{NaOH}=0,3\left(mol\right)\)

=> \(x=CM_{HCl}=\dfrac{0,3}{0,2}=1,5M\)

b) \(n_{NaCl}=n_{NaOH}=0,3\left(mol\right)\)

=> \(CM_{NaCl}=\dfrac{0,3}{0,3+0,2}=0,6M\)