500ml =0,5l

KOH+HCl->KCl+H₂O

0,5-----0,5----0,5---0,5 mol

n HCl =0,5.1=0,5 mol

VKOH =\(\dfrac{0,5}{2}=0,25l=250ml\)

m KCl =0,5.74,5=37,25g

uiii giỏi z 

\(n_{HNO_3}=n_{NO_3^{^-}}=0,2.2=0,4mol\\ n_{H_2SO_4}=n_{SO_4^{2-}}=0,2.1=0,2mol\\ n_{KOH}=x\left(mol\right);V_{ddBase}=v\left(L\right)\\ H^++OH^-->H_2O\\ 0,4+0,4=x+2.0,5.v\\ x+v=0,8\left(I\right)\\ m_{rắn}=62.0,4+96.0,2+39x+137.v.0,5=87\\ 39x+68,5v=43\left(II\right)\\ \Rightarrow x=v=0,4\\ V=1000v=400\left(mL\right)\)

Bài 1:

Ta có: \(n_{HCl}=0,2.0,5=0,1\left(mol\right)\)

BTNT H, có: \(n_{HCl}=2n_{H_2O}\Rightarrow n_{H_2O}=0,05\left(mol\right)\)

Theo ĐL BTKL, có: m _oxit + m_HCl = m_muối + m_H2O

⇒ m_muối = 2,8 + 0,1.36,5 - 0,05.18 = 5,55 (g)

Bài 2:

\(m_{KOH}=200.5,6\%=11,2\left(g\right)\Rightarrow n_{KOH}=\dfrac{11,2}{56}=0,2\left(mol\right)\)

PT: \(2KOH+CuCl_2\rightarrow2KCl+Cu\left(OH\right)_2\)

Theo PT: \(n_{Cu\left(OH\right)_2}=\dfrac{1}{2}n_{KOH}=0,1\left(mol\right)\)

\(\Rightarrow m_{Cu\left(OH\right)_2}=0,1.98=9,8\left(g\right)\)

PTHH : \(Fe_3O_4+4H_2\rightarrow3Fe+4H_2O\)

.............0,05........0,2.......0,15.........

Có : \(\left\{{}\begin{matrix}n_{H_2}=0,2\left(mol\right)\\n_{Fe_3O_4}=0,075\left(mol\right)\end{matrix}\right.\)

- Theo phương pháp ba dòng .

=> Sau phản ứng H2 hết, Fe3O4 còn dư ( dư 0,025 mol )

=> \(m=m_{Fe3o4du}+m_{Fe}=14,2\left(g\right)\)

b, \(Fe+2HCl\rightarrow FeCl_2+H_2\)

...0,15.....0,3.........0,15..............

\(Fe_3O_4+8HCl\rightarrow2FeCl_3+FeCl_2+4H_2O\)

.0,025......0,2..........0,05.........0,025...................

Có : \(V=\dfrac{n}{C_M}=\dfrac{n}{1}=n_{HCl}=0,2+0,3=0,5\left(l\right)\)

Lại có : \(m_M=m_{FeCl2}+m_{FeCl3}=30,35\left(g\right)\)

cho hỏi phương pháp 3 dòng là j thế

a, Ta co pthh

P2O5 + 4H3PO4 \(\rightarrow\) 3H₄P₂O₇ (1)

H₄P₂O₇ + 4 KOH \(\rightarrow\)K₄P₂O7 \(\downarrow\) + 4 H₂O (2)

b, Theo de bai ta co

mct=mH3PO4=\(\dfrac{mdd.C\%}{100\%}=\dfrac{19,60.5\%}{100\%}=0,98\left(g\right)\)

\(\rightarrow\) nH3PO4=\(\dfrac{0,98}{98}=0,01\left(mol\right)\)

Theo de bai ta co

nKOH=CM.V = 1 .0,1 = 0,1 (mol)

Theo pthh 1

nH4P2O7=\(\dfrac{3}{4}nH3PO4=\dfrac{3}{4}.0,01=0,0075\left(mol\right)\)

Theo pthh 2 ta co

nH4P2O5 = \(\dfrac{0,0075}{1}mol< nKOH=\dfrac{0,1}{4}mol\)

\(\Rightarrow\) So mol cua KOH du (tinh theo so mol cua H4P2O7)

6,48 gam chat ran sau khi co can bao gom K4P2O7 va KOH du

Theo pthh 2

nK4P2O7 = nH4P2O7 = 0,0075 mol

\(\Rightarrow\) mK4P2O7 = 0,0075.330=2,475 (g)

mKOH (du) = 6,48-2,475 = 4,005 (g)

Theo pthh 1

nP2O5 = \(\dfrac{1}{4}nH3PO4=\dfrac{1}{4}.0,01=0,0025\left(mol\right)\)

\(\Rightarrow\) mP2O5= 0,0025.142=0,355 g

PT ( 1):theo mk nghĩ thì oxit axit ko tác dụng vs dd axit mà