Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(n_K=\dfrac{3,9}{39}=0,1mol\)
\(2K+2H_2O\rightarrow2KOH+H_2\)
0,1 0,1 0,05 ( mol )
\(m_{KOH}=0,1.56=5,6g\)
\(V_{H_2}=0,05.22,4=1,12l\)
a, \(2K+2H_2O\rightarrow2KOH+H_2\)
b, \(n_K=\dfrac{3,9}{39}=0,1\left(mol\right)\)
Theo PT: \(n_{KOH}=n_K=0,1\left(mol\right)\Rightarrow m_{KOH}=0,1.56=5,6\left(g\right)\)
\(n_K=\dfrac{m}{M}=\dfrac{3,9}{39}=0,1\left(mol\right)\\ PTHH:2K+2H_2O->2KOH+H_2\)
tỉ lệ 2 : 2 : 2 ; 1
n(mol) 0,1---.0,1------>0,1------>0,05
\(m_{KOH}=n\cdot M=0,1\cdot\left(39+16+1\right)=5,6\left(g\right)\)
\(2K+2H_2O\rightarrow2KOH+H_2\)
nK= 7,8 : 39 = 0,2(mol)
nH2 = 0,2 : 2 = 0,1 (mol)
VH2 = 0,1 . 22,4 = 2,24(l)
nKOH = nK = 0,2(mol)
mKOH= 0,2.56 = 11,2(g)
a)
\(PTHH:2Al+6HCl->2AlCl_3+3H_2\)
1,3<---4<-------1,3<---------2
b)
\(n_{H_2\left(dktc\right)}=\dfrac{V}{22,4}=\dfrac{44,8}{22,4}=2\left(mol\right)\)
\(m_{AlCl_3}=n\cdot M=1,3\cdot\left(27+35,5\cdot3\right)=173,55\left(g\right)\)
\(m_{Al}=n\cdot M=1,3\cdot27=35,1\left(g\right)\)
\(n_K=\dfrac{3,8}{39}=\dfrac{19}{195}mol\)
\(n_{H_2O}=\dfrac{101,8}{18}=\dfrac{509}{90}mol\)
\(2K+2H_2O\rightarrow2KOH+H_2\)
19/195 < 509/90 ( mol )
19/195 19/195 19/195 ( mol )
Chất dư là H2O
\(m_{H_2O\left(dư\right)}=\left(\dfrac{509}{90}-\dfrac{19}{195}\right).18\approx100,04g\)
\(m_{KOH}=\dfrac{19}{195}.56\approx5,45g\)
\(n_K=\dfrac{3,9}{39}=0,1\left(mol\right)\\ a,2K+2H_2O\rightarrow2KOH+H_2\\ b,n_{KOH}=n_K=0,1\left(mol\right)\\ C\%_{ddKOH}=\dfrac{0,1.56}{150}.100\%\approx3,733\%\)
\(2K+2H_2O\rightarrow2KOH+H_2\)
\(2mol\) \(2mol\)
\(0,1mol\) \(0,1mol\)
\(n_K=\dfrac{m}{M}=\dfrac{3,9}{39}=0,1\left(mol\right)\)
\(m_{KOH}=n.M=0,1.56=5,6\left(g\right)\)
\(C\text{%}=\dfrac{m_{ct}}{m_{dd}}.100\text{%}=\dfrac{5,6}{150}.100\text{%}=\approx3,73\text{%}\)
a) \(n_{K_2O}=\dfrac{37,6}{94}=0,4\left(mol\right)\)
PTHH: K2O + H2O --> 2KOH
0,4------------>0,8
=> mKOH = 0,8.56 = 44,8 (g)
b) \(C_M=\dfrac{0,8}{0,5}=1,6M\)
c)
PTHH: 4K + O2 --to--> 2K2O
0,2<----0,4
=> VO2 = 0,2.22,4 = 4,48 (l)
nSO2=0,1(mol)
nKOH=0,1(mol)
nKOH/nSO2=1/1
=> P.ứ tạo duy nhất muối axit
a) PTHH: SO2 + KOH -> KHSO3
b) nKHSO3=nKOH=0,1(mol)
=>mKHSO3= 120.0,1=12(g)
Chúc em học tốt!
Ta có: \(\left\{{}\begin{matrix}n_{SO_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\\n_{KOH}=0,1\cdot1=0,1\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\)Tạo muối axit
PTHH: \(SO_2+KOH\rightarrow KHSO_3\)
Theo PTHH: \(n_{KHSO_3}=0,1\left(mol\right)\) \(\Rightarrow m_{KHSO_3}=0,1\cdot136=13,6\left(g\right)\)
nK = \(\dfrac{3,9}{39}=0,1\) mol
Pt: 2K + 2H2O --> 2KOH + H2
0,1 mol------------>0,1 mol-> 0,05 mol
mKOH thu được = 0,1 . 56 = 5,6 (g)
VH2 tạo ra = 0,05 . 22,4 = 1,12 (lít)
nK = \(\dfrac{3,9}{39}\) = 0,1 mol
2K + 2H2O →2KOH + H2
0,1mol →0,1mol→0,05mol
mKOH = 0,1. 56 = 5,6 g
VH2 = 0,05 . 22,4 = 1,12 (l)