Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(n_{Na}=\dfrac{9,2}{23}=0,4\left(mol\right)\)
\(pthh:2Na+2H_2O->2NaOH+H_2\)
0,4 0,4 0,2
=> \(V_{H_2}=0,2.22,4=4,48\left(L\right)\\ m_{NaOH}=0,4.40=16\left(G\right)\)
\(a.2Na+2H_2O\rightarrow2NaOH+H_2\\ b.n_{Na}=\dfrac{9,2}{23}=0,4\left(mol\right)\\ n_{H_2}=\dfrac{1}{2}n_{Na}=0,2\left(mol\right)\\ \Rightarrow V_{H_2}=0,2.22,4=4,48\left(l\right)\\ n_{NaOH}=n_{Na}=0,4\left(mol\right)\\ \Rightarrow m_{NaOH}=0,4.40=16\left(g\right)\\ c.H_2+CuO-^{t^o}\rightarrow Cu+H_2O\\ n_{CuO}=\dfrac{12}{80}=0,15\left(mol\right)\\ LTL:\dfrac{0,2}{1}>\dfrac{0,15}{1}\Rightarrow H_2dưsauphảnứng\\ n_{H_2\left(pứ\right)}=n_{CuO}=0,15\left(mol\right)\\ \Rightarrow n_{H_2\left(dư\right)}=0,2-0,15=0,05\left(mol\right)\\ \Rightarrow m_{H_2\left(Dư\right)}=0,05.2=0,1\left(g\right)\)
\(n_{Na}=\dfrac{6,9}{23}=0,3\left(mol\right)\\ 2Na+2H_2O\rightarrow2NaOH+H_2\\ n_{H_2}=\dfrac{0,3}{2}=0,15\left(mol\right)\\ a,V_{H_2\left(đktc\right)}=0,15.22,4=3,36\left(l\right)\\ b,m_{ddsaup.ứ}=m_{Na}+m_{H_2O}-m_{H_2}=6,9+100-0,15.2=106,6\left(g\right)\)
\(n_{Na}=\dfrac{2.3}{23}=0,1\left(mol\right)\)
PTHH : 2Na + 2H2O -> 2NaOH + H2
0,1 0,1 0,1 0,05
\(m_{NaOH}=0,1.40=4\left(g\right)\)
\(m_{H_2O}=47,8\left(g\right)\)
\(m_{H_2}=0,05.2=0,1\left(g\right)\)
\(m_{dd}=47,8+2,3-0,1=50\left(g\right)\)
\(C\%_{NaOH}=\dfrac{4}{50}.100\%=8\%\)
\(47,8ml=47,8g\)
\(n_{Na}=\dfrac{2,3}{23}=0,1mol\)
\(2Na+2H_2O\rightarrow2NaOH+H_2\)
0,1 0,1 0,05 ( mol )
\(m_{NaOH}=0,1.40=4g\)
\(m_{dd}=2,3+47,8-0,05.2=50g\)
\(C\%_{NaOH}=\dfrac{4}{50}.100=14\%\)
a)PTHH: Na+H2O---> NaOH+H2
b)nNa= \(\dfrac{m}{M}=\dfrac{2,3}{23}=0,1\left(mol\right)\)
=>nNa=nH2=0,1 (mol)
=>VH2=n.22,4=0,1.22,4=2,24(l)
c)nNa=nH2O=nNaOH=0,1 (mol)
=>mH2O=0,1.18=1,8(g)
d)mNaOH=0,1.(23+16+1)=4(g)
Học tốt !
a, \(2Na+2H_2O\rightarrow2NaOH+H_2\)
b, \(n_{Na}=\dfrac{2,3}{23}=0,1\left(mol\right)\)
Theo PT: \(n_{H_2}=\dfrac{1}{2}n_{Na}=0,05\left(mol\right)\Rightarrow V_{H_2}=0,05.22,4=1,12\left(l\right)\)
c, \(n_{H_2O}=n_{Na}=0,1\left(mol\right)\Rightarrow m_{H_2O}=0,1.18=1,8\left(g\right)\)
d, \(n_{NaOH}=n_{Na}=0,1\left(mol\right)\Rightarrow m_{NaOH}=0,1.40=4\left(g\right)\)
\(m_{NaOH}=200.10\%=20\left(g\right)\Rightarrow n_{NaOH}=\dfrac{20}{40}=0,5\left(mol\right)\)
PTHH: 2Na + 2H2O → 2NaOH + H2
Mol: 0,5 0,5 0,5
\(m_{Na}=0,5.23=11,5\left(g\right)\)
\(V_{H_2}=0,25.22,4=5,6\left(l\right)\)
mH2O = 47,8 . 1 = 47,8 (g)
nH2O = 47,8/18
nNa = 2,3/23 = 0,1 (mol)
PTHH: 2Na + 2H2O -> 2NaOH + H2
0,1 ---> 0,1 ---> 0,1 ---> 0,05
mNaOH = 0,1 . 40 = 4 (g)
mH2 = 0,05 . 2 = 0,1 (g)
mdd (sau p/ư) = 47,8 + 2,3 - 0,1 = 50 (g)
C%NaOH = 4/50 = 8%
ĐỀ hỏi nồng độ mol hay nồng độ chất tan vậy, nếu hỏi nồng độ chất tan thì đề lỗi
`Fe + H_2 SO_4 -> FeSO_4 + H_2 ↑`
`0,3` `0,3` `0,3` `0,3` `(mol)`
`n_[Fe] = [ 16,8 ] / 56 = 0,3 (mol)`
`a) m_[dd H_2 SO_4] = [ 0,3 . 98 ] / [ 9,8 ] . 100 = 300 (g)`
`b) V_[H_2] = 0,3 . 22,4 = 6,72 (l)`
`c) C%_[FeSO_4] = [ 0,3 . 152 ] / [ 16,8 + 300 - 0,3 . 2 ] . 100 ~~ 14,42%`
nNa = 9.2/23 = 0.4 (mol)
2Na + 2H2O => 2NaOH + H2
0.4.........................0.4.......0.2
VH2 = 0.2 * 22.4 = 4.48 (l)
mNaOH = 0.4 * 40 = 16 (g)
mdd = 9.2 + 100 - 0.2 * 2 = 108.8 (g)
C% NaOH = 16 / 108.8 * 100% = 14.71%
PTHH: \(Na+H_2O\rightarrow NaOH+\dfrac{1}{2}H_2\uparrow\)
Ta có: \(n_{Na}=\dfrac{9,2}{23}=0,4\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}n_{NaOH}=0,4\left(mol\right)\\n_{H_2}=0,2\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{NaOH}=0,4\cdot40=16\left(g\right)\\V_{H_2}=0,2\cdot22,4=4,48\left(l\right)\\m_{H_2}=0,2\cdot2=0,4\left(g\right)\end{matrix}\right.\)
Mặt khác: \(m_{dd\left(saup/ứ\right)}=m_{Na}+m_{H_2O}-m_{H_2}=108,8\left(g\right)\)
\(\Rightarrow C\%_{NaOH}=\dfrac{16}{108,8}\cdot100\%\approx14,71\%\)