Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(n_{FeCl_3}=0.1\cdot0.1=0.01\left(mol\right)\)
\(n_{NaOH}=0.5\cdot0.1=0.05\left(mol\right)\)
\(FeCl_3+3NaOH\rightarrow Fe\left(OH\right)_3+3NaCl\)
\(1............3\)
\(0.01...........0.05\)
Lập tỉ lệ : \(\dfrac{0.01}{1}< \dfrac{0.05}{3}\Rightarrow NaOHdư\)
Các chất có trong D : \(NaCl:0.03\left(mol\right),NaOH\left(dư\right):0.02\left(mol\right)\)
\(V=0.1+0.5=0.6\left(l\right)\)
\(\left[Na^+\right]=\dfrac{0.03+0.02}{0.06}=\dfrac{1}{12}\left(M\right)\)
\(\left[Cl^-\right]=\dfrac{0.03}{0.06}=0.5\left(M\right)\)
\(\left[OH^-\right]=\dfrac{0.02}{0.6}=\dfrac{1}{30}\left(M\right)\)
\(b.\)
\(m_{Fe\left(OH\right)_3}=0.01\cdot107=1.07\left(g\right)\)
a, \(n_{Ba\left(OH\right)_2}=0,1.0,1=0,01\left(mol\right)=n_{Ba^{2+}}\)
\(\Rightarrow n_{OH^-}=2n_{Ba\left(OH\right)_2}=0,02\left(mol\right)\)
\(n_{NaOH}=0,1.0,1=0,01\left(mol\right)=n_{Na^+}=n_{OH^-}\)
⇒ ΣnOH- = 0,02 + 0,01 = 0,03 (mol)
\(n_{H_2SO_4}=0,4.0,0175=0,007\left(mol\right)=n_{SO_4^{2-}}\)
\(\Rightarrow n_{H^+}=2n_{H_2SO_4}=0,014\left(mol\right)\)
\(H^++OH^-\rightarrow H_2O\)
0,014___0,014 (mol) ⇒ nOH- dư = 0,03 - 0,014 = 0,016 (mol)
\(Ba^{2+}+SO_4^{2-}\rightarrow BaSO_4\)
0,007____0,007_____0,007 (mol) ⇒ nBa2+ dư = 0,01 - 0,007 = 0,003 (mol)
⇒ m = 0,007.233 = 1,631 (g)
\(\left[OH^-\right]=\dfrac{0,016}{0,1+0,4}=0,032\left(M\right)\)
\(\left[Ba^{2+}\right]=\dfrac{0,003}{0,1+0,4}=0,006\left(M\right)\)
\(\left[Na^+\right]=\dfrac{0,01}{0,1+0,4}=0,02\left(M\right)\)
b, pH = 14 - (-log[OH-]) ≃ 12,505
\(n_{Ba^{2+}}=0,1.0,1=0,01\left(mol\right)\)
\(n_{SO_4^{2-}}=0,4.0,0175=7.10 ^{-3}\left(mol\right)\)
\(Ba^{2+}+SO_4^{2-}\rightarrow BaSO_4\downarrow\)
\(\Rightarrow m=m_{BaSO_4}=7.10^{-3}.233=1,631\left(g\right)\)
Ta có:
\(n_{H^+}=0,4.0,0175.2=0,014\left(mol\right)\)
\(n_{OH^-}=0,1.0,1.2+0,1.0,1=0,03\left(mol\right)\)
Trong dung dịch X:
\(n_{OH^-}=0,03-0,014=0,016\left(mol\right)\)\(\Rightarrow\left[OH^-\right]=\dfrac{0,016}{0,1+0,4}=0,032\left(M\right)\)
\(n_{Ba^{2+}}=0,01-7.10^{-3}=3.10^{-3}\left(mol\right)\Rightarrow\left[Ba^{2+}\right]=\dfrac{3.10^{-3}}{0,1+0,4}=6.10^{-3}\left(M\right)\)
\(n_{Na^+}=0,1.0,1=0,01\left(mol\right)\Rightarrow\left[Na^+\right]=0,02\)
\(pOH=-lg\left(0,032\right)\approx1,5\Rightarrow pH=14-1,5=12,5\)
a, \(n_{HCl}=0,2.0,1=0,02\left(mol\right)=n_{H^+}=n_{Cl^-}\)
\(n_{H_2SO_4}=0,2.0,15=0,03\left(mol\right)=n_{SO_4^{2-}}\) \(\Rightarrow n_{H^+}=2n_{H_2SO_4}=0,06\left(mol\right)\)
\(\Rightarrow\Sigma n_{H^+}=0,02+0,06=0,08\left(mol\right)\)
\(n_{Ba\left(OH\right)_2}=0,3.0,05=0,015\left(mol\right)=n_{Ba^{2+}}\)
\(\Rightarrow n_{OH^-}=2n_{Ba\left(OH\right)_2}=0,03\left(mol\right)\)
\(H^++OH^-\rightarrow H_2O\)
0,03___0,03 (mol) ⇒ nH+ dư = 0,05 (mol)
\(Ba^{2+}+SO_4^{2-}\rightarrow BaSO_4\)
0,015___0,015______0,015 (mol) ⇒ nSO42- dư = 0,015 (mol)
⇒ m = mBaSO4 = 0,015.233 = 3,495 (g)
\(\left[Cl^-\right]=\dfrac{0,02}{0,2+0,3}=0,04\left(M\right)\)
\(\left[H^+\right]=\dfrac{0,05}{0,2+0,3}=0,1\left(M\right)\)
\(\left[SO_4^{2-}\right]=\dfrac{0,015}{0,2+0,3}=0,03\left(M\right)\)
b, pH = -log[H+] = 1
a) Ta có: \(n_{Al\left(NO_3\right)_3}=\dfrac{4,26}{213}=0,02\left(mol\right)\) \(\Rightarrow\left\{{}\begin{matrix}n_{Al^+}=0,02\left(mol\right)\\n_{NO_3^-}=0,06\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\left[Al^+\right]=\dfrac{0,02}{0,1}=0,2\left(M\right)\\\left[NO_3^-\right]=\dfrac{0,06}{0,1}=0,6\left(M\right)\end{matrix}\right.\)
b) Ta có: \(\left[Na^+\right]=0,1+0,02\cdot2+0,3=0,304\left(M\right)\)
c) Bạn xem lại đề !!
m NaOH=\(\dfrac{20\cdot\left(100\cdot1.25\right)}{100}=25g\)
=>nNaOH=25/40=0.625 mol
Sau khi thêm 400ml dung dịch NaOH 0,5M :
nNaOH= 0.4*0.5=0.2 mol
NaOH ------------> Na+ + OH-
=>nOH- =nNa+= 0.2+0.625=0.825 mol
=>[Na+] =[OH-] = 0.825/0.5=1.65 M.
\(n_{FeCl_3}=0,1.0,1=0,01mol\)
\(n_{NaOH}=0,4.0,1=0,04mol\)
FeCl3+3NaOH\(\rightarrow\)Fe(OH)3\(\downarrow\)+3NaCl
-Tỉ lệ: \(\dfrac{0,01}{1}< \dfrac{0,04}{3}\)\(\rightarrow\)FeCl3 hết, NaOH dư
\(n_{NaOH\left(pu\right)}=3n_{FeCl_3}=0,03mol\)
\(n_{NaOH\left(dư\right)}=0,04-0,03=0,01mol\)
\(n_{NaCl}=3n_{FeCl_3}=0,03mol\)
-Dung dịch D có NaCl 0,03 mol và NaOH dư 0,01 mol
[Na+]=\(\dfrac{0,03+0,01}{0,1+0,4}=0,08M\)
[Cl-]=\(\dfrac{0,03}{0,5}=0,06M\)
[OH-]=\(\dfrac{0,01}{0,5}=0,02M\)
\(n_{Fe\left(OH\right)_3}=n_{FeCl_3}=0,01mol\)
m=\(m_{Fe\left(OH\right)_3}=0,01.107=1,07g\)