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\(a,PTHH:2Na+2H_2O\rightarrow2NaOH+H_2\\ 2K+2H_2O\rightarrow2KOH+H_2\\ n_{Na}=\dfrac{4,6}{23}=0,2\left(mol\right);n_K=\dfrac{3,9}{39}=0,1\left(mol\right)\\ b,n_{H_2\left(tổng\right)}=\dfrac{1}{2}.\left(n_{Na}+n_K\right)=\dfrac{0,2+0,1}{2}=0,15\left(mol\right)\\ V_{H_2\left(đktc\right)}=0,2.22,4=4,48\left(l\right)\)
\(a,n_{Na}=\dfrac{4,6}{23}=0,2\left(mol\right)\\ n_{H_2O}=\dfrac{5,4}{18}=0,3\left(mol\right)\)
PTHH: 2Na + 2H2O ---> 2NaOH + H2
LTL: 0,2 < 0,3 => H2O dư
Theo pthh: \(\left\{{}\begin{matrix}n_{H_2}=\dfrac{1}{2}n_{Na}=\dfrac{1}{2}.0,2=0,1\left(mol\right)\\n_{NaOH}=n_{H_2}=0,2\left(mol\right)\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}b,V_{H_2}=0,1.22,4=2,24\left(l\right)\\m_{NaOH}=0,2.40=8\left(g\right)\end{matrix}\right.\)
\(2Na+2H_2O\rightarrow2NaOH+H_2\\2K+2H_2O\rightarrow2KOH+H_2\\ n_{Na}=0,1\left(mol\right);n_K=0,05\left(mol\right)\\ n_{H_2}=\dfrac{1}{2}n_{Na}+\dfrac{1}{2}n_K=0,05+0,025=0,075\left(mol\right)\\ \Rightarrow V_{H_2}=0,075.22,4=1,68\left(l\right)\)
Bài 13: nNa= 0,2 mol ; nK= 0,1 mol
2Na + 2H2O → 2NaOH + H2↑
0,2 mol 0,2 mol 0,1 mol
2K + 2H2O → 2KOH + H2↑
0,1 mol 0,1 mol 0,05 mol
a) tổng số mol khí H2 là: nH2= 0,1 + 0,05 = 0,15 mol
→VH2= 0,15 x 22,4 = 3,36 (l)
b) mNaOH= 0,2 x 40= 8 (g) ; mKOH= 0,1 x 56= 5,6 (g)
mdung dịch= mNa + mK + mH2O - mH2 = 4,6 + 3,9 + 91,5 - 0,15x2 = 99,7 (g)
→C%NaOH= 8/99,7 x100%= 8,02%
→C%KOH= 5,6/99,7 x100%= 5,62%
Bài 12:
\(n_{Na}=\dfrac{4,6}{23}=0,2\left(mol\right)\)
\(n_{Ca}=\dfrac{8}{40}=0,2\left(mol\right)\)
PTHH: 2Na + 2H2O --> 2NaOH + H2
0,2----------------------->0,1
Ca + 2H2O --> Ca(OH)2 + H2
0,2---------------------->0,2
=> VH2 = (0,1 + 0,2).24,79 = 7,437(l)
nNa = 4,6 : 23 = 0,2 (mol)
2Na + 2H2O - > 2NaOH + H2
0,2 0,1
nCa = 8 : 40 = 0,2 (mol)
Ca + 2H2O -- > Ca(OH)2 + H2
0,2 0,2
nH2 = 0,1 + 0,2 = 0,3 (mol)
VH2 = 0,3 . 24,79 = 7,437 (l)
`CuO+ H_2 -> Cu+ H_2O`
`0,03 ----0,03` mol
`Fe_2O_3+ 3H_2 ->2Fe + 3H_2O`
`0,1-------0,3` mol
`n_(CuO) = 2,4/80 =0,03` mol
`n_(Fe_2O_3)=16/160 =0,1` mol
`=> V_(H_2)=(0,3+0,03).22,4=7,392 l`
có đk nhiệt nha em, p.ứ nay nung nóng mà
2Na+2H2O->2NaOH+H2
0,2-----0,2----0,2----------0,1
n Na=0,2 mol
=>Quỳ chuyển màu xanh
VH2=0,1.22,4=2,24l
2Na+2H2O->2NaOH+H2
n H2O=0,4 mol
=>H2O dư
=>m dư=0,2.18=3,6g
a) QT chuyển xanh
\(n_{Na}=\dfrac{4,6}{23}=0,2\left(mol\right)\\
pthh:Na+H_2O\rightarrow NaOH+\dfrac{1}{2}H_2\)
0,2 0,1
\(V_{H_2}=0,1.22,4=2,24\left(l\right)\\
n_{H_2O}=\dfrac{7,2}{18}=0,4\left(mol\right)\\
LTL:\dfrac{0,2}{1}< \dfrac{0,4}{1}\)
=> H2O dư
\(n_{H_2O\left(p\text{ư}\right)}=n_{Na}=0,2\left(mol\right)\\
m_{H_2O\left(d\right)}=\left(0,4-0,2\right).18=3,6\left(g\right)\)
$1)$
$4Na+O_2\xrightarrow{t^o}2Na_2O$
$3Fe+2O_2\xrightarrow{t^o}Fe_3O_4$
$2Cu+O_2\xrightarrow{t^o}2CuO$
$2Na+2HCl\to 2NaCl+H_2$
$Fe+2HCl\to FeCl_2+H_2$
$2Na+2H_2O\to 2NaOH+H_2$
$2)$
$n_{O_2}=0,225(mol)$
$n_{H_2(TN_2)}=0,25(mol)$
$n_{H_2(TN_3)}=0,1(mol)$
Theo PT: $\begin{cases} n_{Na}=2n_{H_2(TN_3)}=0,2(mol)\\ 0,5n_{Na}+n_{Fe}=n_{H_2(TN_2)}=0,25(mol)\\ 0,25n_{Na}+\dfrac{2}{3}n_{Fe}+0,5n_{Cu}=n_{O_2}=0,225(mol) \end{cases}$
$\to\begin{cases} n_{Na}=0,2(mol)\\ n_{Fe}=0,15(mol)\\ n_{Cu}=0,15(mol) \end{cases}$
$\to \begin{cases} \%n_{Na}=\dfrac{0,2}{0,2+0,15+0,15}.100\%=40\%\\ \%n_{Fe}=\%n_{Cu}=\dfrac{0,15}{0,2+0,15+0,15}.100\%=30\% \end{cases}$
$\to m=0,2.23+0,15.56+0,15.64=22,6(g)$
\(n_{Na}=\dfrac{4,6}{23}=0,2\left(mol\right)\)
\(n_{Ca}=\dfrac{1,2}{40}=0,03\left(mol\right)\)
PT: \(2Na+2H_2O\rightarrow2NaOH+H_2\)
\(Ca+2H_2O\rightarrow Ca\left(OH\right)_2+H_2\)
Theo PT: \(n_{H_2}=\dfrac{1}{2}n_{Na}+n_{Ca}=0,13\left(mol\right)\Rightarrow V_{H_2}=0,13.22,4=2,912\left(l\right)\)