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2)
nKOH = 0.15*2=0.3 mol
nHCl = 0.25*3=0.75 mol
KOH + HCl --> KCl + H2O
Bđ: 0.3____0.45
Pư : 0.3____0.3____0.3
Kt: 0______0.15___0.3
DD sau phản ứng : 0.15 mol HCl dư , 0.3 mol KCl
CM H+= 0.15/0.25=0.6M
CM Cl- = 0.15/0.25=0.6 M
CM K+= 0.3/(0.15+0.25)=0.75M
CM Cl-= 0.3/(0.15+0.25)= 0.75M
a)
$BaCl_2 + Na_2SO_4 \to BaSO_4 + 2NaCl$
$n_{BaCl_2} = 0,01 = n_{Na_2SO_4} = 0,01 \Rightarrow $ Vừa đủ
$n_{BaSO_4} = n_{Na_2SO_4} = 0,01(mol)$
$m_{BaSO_4} = 0,01.233 = 0,233(gam)$
b)
$n_{NaCl} = 2n_{Na_2SO_4} = 0,02(mol)$
$V_{dd} = 0,1 + 0,2 = 0,3(lít)$
$C_{M_{NaCl}} = \dfrac{0,02}{0,3} = 0,067M$
c)
$[Na^+] = [Cl^-] = C_{M_{NaCl}} = 0,067M$
\(n_{BaCl_2}=0.1\cdot0.1=0.01\left(mol\right)\)
\(n_{Na_2SO_4}=0.2\cdot0.05=0.01\left(mol\right)\)
\(BaCl_2+Na_2SO_4\rightarrow BaSO_4+2NaCl\)
\(0.01..........0.01............0.01..............0.02\)
\(m_{BaSO_4}=0.01\cdot233=2.33\left(g\right)\)
\(C_{M_{NaCl}}=\dfrac{0.01}{0.1+0.2}=0.03\left(M\right)\)
\(\left[Na^+\right]=\left[Cl^-\right]=0.03\left(M\right)\)
a, \(\left\{{}\begin{matrix}n_{Ba^{2+}}=4.10^{-3}\left(mol\right)\\n_{Na^+}=3.10^{-3}\left(mol\right)\\n_{OH^-}=0,011\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\left[Ba^{2+}\right]=\dfrac{4.10^{-3}}{0,2+0,3}=0,008M\\\left[Na^+\right]=\dfrac{3.10^{-3}}{0,2+0,3}=0,006M\\\left[OH^-\right]=\dfrac{0,011}{0,2+0,3}=0,022M\end{matrix}\right.\)
b, Để trung hòa dung dịch A thì:
\(n_{H^+}=n_{OH^-}\)
\(\Leftrightarrow0,01.V_{ddHCl}=\left(0,02.2+0,01\right).0,2\)
\(\Leftrightarrow V_{ddHCl}=1\left(l\right)\)
a)\(BaCl_2\rightarrow Ba^{2+}+2Cl^-\)
0,05 0,05 0,1
\(\left[Ba^{2+}\right]=0,05M\)
\(\left[Cl^-\right]=0,1M\)
b)\(HCl\rightarrow H^++Cl^-\)
0,1 0,1 0,1
\(\left[H^+\right]=0,1M\)
\(\left[Cl^-\right]=0,1M\)
a)
[K+]=[OH-]=0,02M
b)
[Ba2+]=0,015M
[Cl-]=0,03M
c)
[H+]=0,05M
[Cl-]=0,05M
d)
[NH4+]=0,02M
[SO42-]=0,01M