a)

$m_{dd} = 16 + 234 = 250(gam)$
$V_{dd} = \dfrac{250}{1,05} = 238(ml)$

$n_{NaOH} = \dfrac{16}{40} = 0,4(mol)$

Suy ra :

$C\%_{NaOH} = \dfrac{16}{250} = 6,4\%$

$C_{M_{NaOH}} = \dfrac{0,4}{0,238} = 1,68M$

b)

$C\%_{NaOH} = \dfrac{16+10}{250+10}.100\% = 10\%$

Quy hh về FeO : a mol và Fe2O3: b mol (trong từng phần)
Phần 1: mFe2O3=8,8g --> nFe2O3=0,055 mol -->a/2 +b=0,055
Phần 2: nKMnO4=0,01 mol -->n Fe2+=0,05=nFeO=a
-->b=0,03 mol
m=16,8 g ; nH2SO4=nO=0,28 mol -->V=0,56l

a) \(X+H_2O\rightarrow XOH+\dfrac{1}{2}H_2\)

\(m_{ddA}=m_X+m_{H_2O}-m_{H_2}\)

=> \(m_{H_2}=0,3\left(g\right)\Rightarrow n_{H_2}=0,15\left(mol\right)\)

Theo PT \(n_X=2n_{H_2}=0,3\left(mol\right)\)

=> \(M_X=\dfrac{11,7}{0,3}=39\)

Vậy X là Kali

b) \(n_{KOH}=n_K=0,3\left(mol\right)\)

\(C\%_{KOH}=\dfrac{0,3.56}{132}.100=12,73\%\)

a)Gọi hóa trị của M trong muối clorua là n
Gọi hóa trị của M trong muối nitrat là m

2M + 2nHCl -----> 2MCln + nH2
_1___________________\(\dfrac{n}{2}\)
3M + 4mHNO3 -----> 3M(NO3)m + mNO + 2mH2O
_1_________________________\(\dfrac{m}{3}\)_

Ta có: VH2 = VNO => nH2 = nNO
\(\dfrac{m}{3}=\dfrac{n}{2}\) --> \(\dfrac{n}{m}=\dfrac{2}{3}\) => n = 2; m = 3

Vậy hóa trị của M trong muối clorua < hóa trị của M trong muối nitrat

b,
mM(NO3)m = 1,905m.MCln
M + 62m = 1,905x(M + 35,5n)
<=> 0,905M + 67,6275n = 62m
<=> M = \(\dfrac{62m-67,6275n}{0,905}\)
Thay n = 2; m = 3 vào ta được
M = 56 (Fe)

a)

2M + 2nHCl --> 2MCl_n + nH₂

3M + 4mHNO₃ --> 3M(NO₃)_m + mNO + 2mH₂O

b) 

\(n_M=\dfrac{a}{M_M}\left(mol\right)\)

PTHH: 2M + 2nHCl --> 2MCl_n + nH₂

          \(\dfrac{a}{M_M}\)--------------------->\(\dfrac{an}{2.M_M}\)

           3M + 4mHNO₃ --> 3M(NO₃)_m + mNO + 2mH₂O

           \(\dfrac{a}{M_M}\)-------------------------->\(\dfrac{am}{3.M_M}\)

=> \(\dfrac{an}{2.M_M}=\dfrac{am}{3.M_M}\)

=> \(\dfrac{n}{m}=\dfrac{2}{3}< 1\)

=> n < m

c) 

Có: n = 2; m = 3

a)

2M + 2nHCl --> 2MCl_n + nH₂

3M + 4mHNO₃ --> 3M(NO₃)_m + mNO + 2mH₂O

b)

\(n_M=\dfrac{a}{M_M}\left(mol\right)\)

PTHH: 2M + 2nHCl --> 2MCl_n + nH₂

           \(\dfrac{a}{M_M}\)------------------->\(\dfrac{an}{2.M_M}\)

            3M + 4mHNO₃ --> 3M(NO₃)_m + mNO + 2mH₂O

          \(\dfrac{a}{M_M}\)---------------------------->\(\dfrac{am}{3.M_M}\)

=> \(\dfrac{an}{2.M_M}=\dfrac{am}{3.M_M}\)

=> \(\dfrac{n}{m}=\dfrac{2}{3}\)

=> n < m

c) Chọn n = 2; m = 3

PTHH: M + 2HCl --> MCl₂ + H₂

         \(\dfrac{a}{M_M}\)--------->\(\dfrac{a}{M_M}\)

            M + 4HNO₃ --> M(NO₃)₃ + NO + 2H₂O

            \(\dfrac{a}{M_M}\)----------->\(\dfrac{a}{M_M}\)

=> \(\left\{{}\begin{matrix}m_{MCl_2}=\dfrac{a}{M_M}\left(M_M+71\right)\\m_{M\left(NO_3\right)_3}=\dfrac{a}{M_M}\left(M_M+186\right)\end{matrix}\right.\)

=> \(\dfrac{M_M+186}{M_M+71}=1,905\)

=> M_M = 56 (g/mol)

=> M là Fe

nNa2O= 12,4/62=0,2(mol)

a) PTHH: Na2O + H2O -> 2 NaOH

0,2_________0,2____0,4(mol)

b) VddNaOH=2(l)

=>CMddNaOH=0,4/0,2=2(M)

Chúc em học tốt!

E cảm ơn ạ

\(n_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right);n_{HCl}=0,2.2,5=0,5\left(mol\right)\\ Fe+2HCl\rightarrow FeCl_2+H_2\\ Vì:\dfrac{0,2}{1}< \dfrac{0,5}{2}\Rightarrow HCl.dư\\ n_{H_2}=n_{FeCl_2}=n_{Fe}=0,2\left(mol\right);n_{HCl\left(dư\right)}=0,5-0,2.2=0,1\left(mol\right)\\ a,V_{H_2\left(đktc\right)}=0,2.22,4=4,48\left(l\right)\\ b,V_{ddsau}=V_{ddHCl}=0,2\left(l\right)\\ C_{MddFeCl_2}=\dfrac{0,2}{0,2}=1\left(M\right);C_{MddHCl\left(dư\right)}=\dfrac{0,1}{0,2}=0,5\left(M\right)\)