Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Đặt x, y là số mol HCl và H 2 SO 4 trong 40 ml dung dịch A.
HCl + NaOH → NaCl + H 2 O
H 2 SO 4 + 2NaOH → Na 2 SO 4 + 2 H 2 O
Số mol NaOH: x + 2y = 1.60/1000 = 0,06 mol (1)
Khối lượng 2 muối : 58,5x + 142y = 3,76 (2)
Từ (1), (2), giải ra : x = 0,04 ; y = 0,01.
C M HCl = 0,04/0,04 = 1(mol/l)
C M H 2 SO 4 = 0,01/0,04 = 0,25 (mol/l)
\(n_{CaCO_3}=\dfrac{40}{100}=0,4\left(mol\right)\\ CaCO_3+2HCl\rightarrow CaCl_2+CO_2+H_2O\\ CO_2+2NaOH\rightarrow Na_2CO_3+H_2O\\ n_{CO_2}=n_{CaCO_3}=0,4\left(mol\right)\\ \Rightarrow n_{NaOH}=0,4.2=0,8\left(mol\right)\\ m_{NaOH}=0,8.40=32\left(g\right)\\ \Rightarrow m_{ddNaOH}=\dfrac{32.100}{25}=128\left(g\right)\\ \Rightarrow V_{ddNaOH}=\dfrac{m_{ddNaOH}}{D_{ddNaOH}}=\dfrac{128}{1,28}=100\left(ml\right)=0,1\left(l\right)\)
Đáp án D
nNaOH = 0,2.1,5=0,3 (mol)
NaOH + HCl → NaCl + H2O
0,3 → 0,3 (mol)
$200ml=0,2l$
$n_{NaOH}=0,2.1,5=0,3(mol)$
$H^++OH^-\to H_2O$
$\to n_{H^+}=n_{HCl}=n_{OH^-}=n_{NaOH}=0,3(mol)$
$\to V_{dd\,HCl}=\dfrac{0,3}{0,4}=0,75(l)=750(ml)$
1)
- TN1:
\(n_{AgCl}=\dfrac{35,875}{143,5}=0,25\left(mol\right)\)
PTHH: AgNO3 + HCl --> AgCl + HNO3
0,25<--0,25
TN2:
nNaOH = 0,5.0,3 = 0,15 (mol)
PTHH: NaOH + HCl --> NaCl + H2O
0,15--->0,15
\(n_{HCl\left(dd.C\right)}=0,25+0,15\) = 0,4 (mol)
=> \(C_{M\left(dd.C\right)}=\dfrac{0,4}{2}=0,2M\)
2)
Có \(\left\{{}\begin{matrix}C_{M\left(A\right)}=\dfrac{0,25}{V}M\\C_{M\left(B\right)}=\dfrac{0,15}{V^,}M\end{matrix}\right.\)
nHCl(A) = \(\dfrac{0,025}{V}\left(mol\right)\)
PTHH: Fe + 2HCl --> FeCl2 + H2
\(\dfrac{0,025}{V}\)------>\(\dfrac{0,0125}{V}\)
nHCl(B) = \(\dfrac{0,015}{V^,}\) (mol)
PTHH: Fe + 2HCl --> FeCl2 + H2
\(\dfrac{0,015}{V^,}\)-------->\(\dfrac{0,0075}{V^,}\)
TH1: \(\dfrac{0,0125}{V}=\dfrac{0,0075}{V^,}+0,02\)
Mà V + V' = 2 (l)
=> \(\left[{}\begin{matrix}V=1,5;V^,=0,5\left(KTM\right)\\V=0,5;V^,=1,5\left(TM\right)\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}C_{M\left(A\right)}=\dfrac{0,25}{0,5}=0,5M\\C_{M\left(B\right)}=\dfrac{0,15}{1,5}=0,1M\end{matrix}\right.\)
TH2: \(\dfrac{0,0125}{V}+0,02=\dfrac{0,0075}{V^,}\)
=> \(\left[{}\begin{matrix}V=\dfrac{1+\sqrt{6}}{2};V^,=\dfrac{3-\sqrt{6}}{2}\\V=\dfrac{1-\sqrt{6}}{2}\left(L\right)\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}C_{M\left(A\right)}=\dfrac{0,25}{\dfrac{1+\sqrt{6}}{2}}=\dfrac{-1+\sqrt{6}}{10}M\\C_{M\left(B\right)}=\dfrac{0,15}{\dfrac{3-\sqrt{6}}{2}}=\dfrac{3+\sqrt{6}}{10}M\end{matrix}\right.\)
Có \(m_M=m_{Na}+m_{Cl}+m_{SO_4}=1,38+m_{Cl}+m_{SO_4}=3,76\)
\(\Rightarrow m_{Cl}+m_{SO_4}=2,38\)
\(\Rightarrow35,5n_{HCl}+96n_{H_2SO_4}=2,38\)
Lại có : \(n_O=n_{NaOH}=0,06\left(mol\right)\)
BtH : \(n_H=2n_{H_2O}=2n_O=n_{H\left(NaOH\right)}+n_{H\left(HCl\right)}+2n_{H\left(H_2SO_4\right)}\)
\(\Rightarrow n_{HCl}+2n_{H_2SO_4}=0,06\)
\(\Rightarrow n_{HCl}=0,04\left(mol\right)\)
\(\Rightarrow C_{MHCl}=1M\)
\(H_3PO_4+3NaOH->Na_3PO_4+3H_2O\\ V_{NaOH}=\dfrac{0,05\cdot0,5\cdot3}{1}=0,075L=75mL\)