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nNa=4,6/23=0,2(mol)
PTHH: Na + H2O -> NaOH + 1/2 H2
0,2_______________0,2____0,1(mol)
mddNaOH=4,6+100-0,1.2=104,4(g)
mNaOH=0,2.40=8(g)
=>C%ddNaOH= (8/104,4).100=7,663%
=> Chọn B (gần nhất)
K + H2O -------> KOH + 1/2 H2
nK = 5,85/39=0,15 (mol)
Theo PT : nKOH=nK = 0,15 (mol)
=> CM KOH = n/V = 0,15/0,1=1,5M
=> Chọn C
Số mol của kali
nK = \(\dfrac{m_K}{M_K}=\dfrac{5,85}{39}=0,15\left(mol\right)\)
Pt : 2K + 2H2O → 2KOH + H2\(|\)
2 2 2 1
0,15 0,15
Số mol của dung dịch kali hidroxit
nKOH= \(\dfrac{0,15.2}{2}=0,15\left(mol\right)\)
Nồng độ mol của dung dịch kali hidroxit
CMKOH = \(\dfrac{0,15}{0,1}=1,5\left(M\right)\)
⇒ Chọn câu : C
Chúc bạn học tốt
\(Na+H_2O \to NaOH + \frac{1}{2}H_2\\ n_{Na}=\frac{4,6}{23}=0,2(mol)\\ n_{NaOH}=n_{Na}=0,2(mol)\\ CM_{NaOH}=\frac{0,2}{0,1}=2M\)
\(n_{K_2O}=\dfrac{6,58}{94}=0,07\left(mol\right);n_K=\dfrac{5,85}{39}=0,15\left(mol\right)\)
PTHH: \(4K+O_2\xrightarrow[]{t^o}2K_2O\)
0,14<--------0,07
\(\Rightarrow H=\dfrac{0,14}{0,15}.100\%=93,33\%\)
PTHH: \(K_2O+H_2O\rightarrow2KOH\)
0,07------------->0,14
\(\Rightarrow C_{M\left(KOH\right)}=\dfrac{0,14}{0,2}=0,7M\)
\(K+H_2O\rightarrow KOH+\dfrac{1}{2}H_2\)
\(n_K=2n_{H_2}=0,2\left(mol\right)\Rightarrow m_K=7,8\left(g\right)\)
=> \(m_{K_2O}=17,2-7,8=9,4\Rightarrow n_{K_2O}=0,1\left(mol\right)\)
\(K_2O+H_2O\rightarrow2KOH\)
\(\Sigma n_{KOH}=0,2+0,1.2=0,4\left(mol\right)\)
\(m_{ddsaupu}=17,2+600-0,1.2=617\left(g\right)\)
=> \(C\%_{KOH}=\dfrac{0,4.56}{712}.100=3,15\%\)
mH2O=71,8.1=71,8(g)
nK2O=28,2/94=0,3(mol)
PTHH: K2O + H2O -> 2 KOH
0,3____________0,6(mol)
mKOH= 0,6.56=33,6(g)
mddKOH= mK2O + mH2O= 28,2+71,8=100(g)
=>C%ddB=C%ddKOH=(33,6/100).100=33,6%
1)
$n_{Na_2O} = \dfrac{6,2}{62} = 0,1(mol)$
$Na_2O + H_2O \to 2NaOH$
$n_{NaOH} = 2n_{Na_2O} = 0,2(mol)$
$m_{dd} = 6,2 + 193,8 = 200(gam) \Rightarrow C\%_{NaOH} = \dfrac{0,2.40}{200}.100\% = 4\%$
2)
$n_{K_2O} = \dfrac{23,5}{94} = 0,25(mol)$
$K_2O + H_2O \to 2KOH$
$n_{KOH} = 2n_{K_2O} = 0,5(mol) \Rightarrow C_{M_{KOH}} = \dfrac{0,5}{0,5} = 1M$
3) $n_{Na_2O} = \dfrac{12,4}{62} = 0,2(mol)$
$Na_2O + H_2O \to 2NaOH$
$n_{NaOH} = 2n_{Na_2O} = 0,4(mol)$
$C_{M_{NaOH}} = \dfrac{0,4}{0,5} =0,8M$
4)
$Na_2SO_3 + 2HCl \to 2NaCl +S O_2 + H_2O$
Theo PTHH :
$n_{SO_2} = n_{Na_2SO_3} = \dfrac{12,6}{126} = 0,1(mol)$
$V_{SO_2} = 0,1.22,4 = 2,24(lít)$
5) $n_{CaO} = \dfrac{5,6}{56} = 0,1(mol)$
$CaO + 2HCl \to CaCl_2 + H_2O$
Theo PTHH :
$n_{HCl} = 2n_{CaO} = 0,2(mol) \Rightarrow m_{dd\ HCl} = \dfrac{0,2.36,5}{14,6\%} = 50(gam)$
\(n_{AlCl3}=\dfrac{26,7}{133,5}=0,2\left(mol\right)\)
\(n_{Al\left(OH\right)3}=\dfrac{7,8}{78}=0,1\left(mol\right)\)
Vì \(n_{Al\left(OH\right)3}< n_{AlCl3}\) nên có 2 gtr \(n_{KOH}\) thỏa mãn
PTHH : \(AlCl_3+3KOH-->Al\left(OH\right)_3+3KCl\) (1)
\(Al\left(OH\right)_3+KOH-->KAlO_2+2H_2O\) (2)
TH1 : KOH thiếu => Chỉ xảy ra pứ (1)
\(n_{KOH}=3n_{Al\left(OH\right)3}=0,3\left(mol\right)\)
=> \(C\%KOH=\dfrac{56.0,3}{400}.100\%=4,2\%=a\%\)
=> a = 4,2
TH2 : KOH dư => Xảy ra pứ (1) và (2)
Có : \(n_{KOH}=3n_{AlCl3}+n_{Al\left(OH\right)3tan}\)
\(=3n_{AlCl3}+\left(n_{Al\left(OH\right)3sinhra}-n_{Al\left(OH\right)3thuduoc}\right)\)
\(=3n_{AlCl3}+\left(n_{AlCl3}-n_{Al\left(OH\right)3thuduoc}\right)\)
\(=4n_{AlCl3}-n_{Al\left(OH\right)3thuduoc}\)
\(=4.0,2-0,1=0,7\left(mol\right)\)
=> \(C\%KOH=\dfrac{0,7.56}{400}.100\%=9,8\%=a\%\)
=> a = 9,8
\(n_K=\frac{5,85}{15}=0,15(mol)\\ K+H_2O \to KOH +\frac{1}{2}H_2\\ n_{KOH}=n_K=0,15(mol)\\ n_{H_2}=\frac{1}{2}.n_K=\frac{1}{2}.0,15=0,075(mol)\\ m_{dd}=5,85+100-(0,075.2)=105,7(g)\\ C\%=\frac{0,15.56}{105,7}.100=7,95\%\)