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PTHH: \(Na+H_2O\rightarrow NaOH+\dfrac{1}{2}H_2\uparrow\) (1)
\(Na_2O+H_2O\rightarrow2NaOH\) (2)
a) Ta có: \(n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\) \(\Rightarrow n_{Na}=0,2\left(mol\right)\)
\(\Rightarrow m_{Na}=0,2\cdot23=4,6\left(g\right)\) \(\Rightarrow m_{Na_2O}=6,2\left(g\right)\)
b) Theo các PTHH: \(\left\{{}\begin{matrix}n_{NaOH\left(1\right)}=n_{Na}=0,2\left(mol\right)\\n_{NaOH\left(2\right)}=2n_{Na_2O}=2\cdot\dfrac{6,2}{62}=0,2\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow n_{NaOH}=0,4\left(mol\right)\) \(\Rightarrow m_{NaOH}=0,4\cdot40=16\left(g\right)\)
c) Theo các PTHH: \(n_{H_2O}=n_{Na}+n_{Na_2O}=0,3\left(mol\right)\)
\(\Rightarrow m_{H_2O}=0,3\cdot18=5,4\left(g\right)\)
a) Gọi số mol Na, Ca là a, b (mol)
=> 23a + 40b = 8,3 (1)
PTHH: 2Na + 2H2O --> 2NaOH + H2
a--------------->a------>0,5a
Ca + 2H2O --> Ca(OH)2 + H2
b--------------->b--------->b
=> \(n_{H_2}=0,5a+b=\dfrac{4,48}{22,4}=0,2\left(mol\right)\) (2)
(1)(2) => a = 0,1; b = 0,15
=> \(\left\{{}\begin{matrix}\%m_{Na}=\dfrac{0,1.23}{8,3}.100\%=27,71\%\\\%m_{Ca}=\dfrac{0,15.40}{8,3}.100\%=72,29\%\end{matrix}\right.\)
b) \(\left\{{}\begin{matrix}m_{NaOH}=0,1.40=4\left(g\right)\\m_{Ca\left(OH\right)_2}=0,15.74=11,1\left(g\right)\end{matrix}\right.\)
\(n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\\ 2Na+2H_2O\rightarrow2NaOH+H_2\\ Ca+2H_2O\rightarrow Ca\left(OH\right)_2+H_2\\ Đặt:n_{Na}=a\left(mol\right);n_{Ca}=b\left(mol\right)\left(a,b>0\right)\\ \Rightarrow\left\{{}\begin{matrix}23a+40b=8,3\\0,5a+b=0,2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,1\\b=0,15\end{matrix}\right.\\ b,\Rightarrow\%m_{Ca}=\dfrac{0,15.40}{8,3}.100\approx72,289\%\\ \Rightarrow\%m_{Na}\approx27,711\%\\ b,n_{NaOH}=a=0,1\left(mol\right)\\ n_{Ca\left(OH\right)_2}=b=0,15\left(mol\right)\\ m_{bazo}=m_{NaOH}+m_{Ca\left(OH\right)_2}=40.0,1+74.0,15=15,1\left(g\right)\)
2Na+2H2O->2NaOH+H2
x------------------------------0,5x
Ca+2H2O->Ca(OH)2+H2
y-------------------------------y
Ta có :
\(\left\{{}\begin{matrix}23x+40y=8,3\\0,5x+y=0,2\end{matrix}\right.=>\left\{{}\begin{matrix}x=0,1\\y=0,15\end{matrix}\right.\)
=>%mNa=\(\dfrac{0,1.23}{8,3}.100=27,71\%\)
=>%mCa=72,29%
b)m bazo=0,1.40+0,15.74=15,1g
\(a) Ca + 2H_2O \to Ca(OH)_2 + H_2\\ CaO + H_2O \to Ca(OH)_2\\ n_{Ca} = n_{H_2} = \dfrac{3,36}{22,4} = 0,15(mol)\\ m_{Ca} = 0,15.40 = 6(gam)\\ \Rightarrow m_{CaO} = 17,2 - 6 = 11,2(gam)\\ b) n_{Ca(OH)_2} = n_{Ca} + n_{CaO} = 0,15 + \dfrac{11,2}{56} = 0,35(mol)\\ m_{Ca(OH)_2} = 0,35.74 = 25,9(gam)\)
\(n_C = a(mol) ; n_S = b(mol)\\ \Rightarrow 12a + 32b = 5,6(1)\\ C + O_2 \xrightarrow{t^o} CO_2\\ S + O_2 \xrightarrow{t^o} SO_2\\ n_{O_2} = n_C + n_S = a + b = \dfrac{9,6}{32} = 0,3(2)\\ (1)(2)\Rightarrow a = 0,2 ; b = 0,1\\ \%m_C = \dfrac{0,2.12}{5,6}.100\% =42,86\%\\ \%m_S = 100\%-42,86\% = 57,14\%\)
\(n_{CO_2} = n_C = 0,2(mol)\\ n_{SO_2} = n_S = 0,1(mol)\\ \%V_{CO_2} = \dfrac{0,2}{0,2 + 0,1}.100\% = 66,67\%\\ \%V_{SO_2} = 100\%-66,67\% = 33,33\%\\ m_{hh\ sau\ pư} = m_C + m_S + m_{O_2} = 5,6 + 9,6 = 15,2(gam)\\ \%m_{CO_2} = \dfrac{0,2.44}{15,2}.100\% = 57,89\%\\ \%m_{SO_2} = 100\% -57,89\% = 42,11\%\)
Ca+ 2H2O -> Ca(OH)2+ H2
nH2= nCa= 0,14 mol
=> mCa= 5,6g
=> mFe= 6,2-5,6= 0,6g
H2 + O -> H2O
=> Y có 0,14 mol O
nFe2O3= 0,02 mol
=> 0,02 mol Fe2O3 có 0,04 mol Fe và 0,06 mol O
Tổng mol Fe sau phản ứng là \(\dfrac{5,6}{56}\)= 0,1 mol
=> FexOy có 0,06 mol Fe và 0,08 mol O
nFe : nO= 0,06 : 0,08= 3 : 4
=> FexOy là Fe3O4
a= 0,06.56+ 0,08.16= 4,64g
\(n_{H_2}=\dfrac{3,136}{22,4}=0,14mol\)
Gọi \(\left\{{}\begin{matrix}n_{Ca}=x\\n_{Na}=y\end{matrix}\right.\)
\(Ca+2H_2O\rightarrow Ca\left(OH\right)_2+H_2\)
x x ( mol )
\(2Na+2H_2O\rightarrow2NaOH+H_2\)
y 1/2 y ( mol )
Ta có:
\(\left\{{}\begin{matrix}40x+23y=6,2\\x+\dfrac{1}{2}y=0,14\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=0,04\\y=0,2mol\end{matrix}\right.\)
\(\rightarrow\left\{{}\begin{matrix}m_{Ca}=0,04.40=1,6g\\m_{Na}=0,2.23=4,6g\end{matrix}\right.\)
\(n_{Fe_2O_3}=\dfrac{3,2}{160}=0,02mol\)
\(Fe_2O_3+3H_2\rightarrow\left(t^o\right)2Fe+3H_2O\)
0,02 0,06 0,04 0,06 ( mol )
\(m_{Fe}=0,04.56=2,24g\)
\(\rightarrow m_{H_2\left(tdFe_xO_y\right)}=0,14-0,06=0,08mol\)
\(n_{Fe\left(tdFe_xO_y\right)}=\dfrac{5,6-0,04.56}{56}=0,06mol\)
\(Fe_xO_y+yH_2\rightarrow\left(t^o\right)xFe+yH_2O\)
0,08 0,06 ( mol )
\(\Rightarrow x:y=0,06:0,08=3:4\)
\(\Rightarrow CTHH:Fe_3O_4\)
a, PTHH : 2K + 2H2O -> 2KOH + H2 (1)
2Na + 2H2O -> 2NaOH + H2 (2)
Khí A là H2
Dung dịch B : KOH , NaOH
mNa =8,5 . 54,12% = 4,6 (g) -(chỗ này có làm tròn xíu lát tính cho dễ )
=>nNa=4,6 /23 = 0,2(mol)
Theo (2) nH2= 1/2 nNa = 0,1 (mol)
Theo (2), nNaOH= nNa = 0,2 (mol)=> mNaOH= 0,2 . 40 = 8(g)
=> mK = 8,5 - 4,6= 3,9 (g)
=>nK = 3,9 / 39 = 0,1 (mol)
Theo (1) , nH2= 1/2 nK = 0,05 (mol)
Theo (1) ,n KOH= nK = 0,1 (mol)=> mKOH = 0,1 . 56=5,6 (g)
\(\Sigma\)nH2= 0,1 + 0,05 = 0,15 (mol)=> VA= 3,36 (l)
a) mNa=4,6g;mK=3,9g ( cái này bạn bik)
pt:\(2Na+2H_2O->2NaOH+H_2\uparrow\)
0,2 ---------------------> 0,2--------->0,1 (mol)
\(2K+2H_2O->2KOH+H_2\uparrow\)
01,------------------->0,1-------------->0,05 ( mol)
VH2= (0,1+0,05).22,4=3,36lit
mNaOH=0,2.40=8g
mKOH=0,156=5,6g
b làm tương tự câu a
\(n_{H_2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\)
PTHH: \(2Na+2H_2O\rightarrow2NaOH+H_2\)
\(2K+H_2O\rightarrow2KOH+H_2\)
Gọi số mol của \(Na\) là : a
Gọi số mol của \(K\) là : b
Ta có:
\(\left\{{}\begin{matrix}23a+39b=14,7\\0,5\left(a+b\right)=0,25\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}a=0,3\\b=0,2\end{matrix}\right.\)
\(m_{Na}=23.0,3=6,9\left(g\right)\)
\(m_K=39.0,2=7,8\left(g\right)\)
PTHH: \(2Na+2H_2O\rightarrow2NaOH+H_2\uparrow\) (1)
\(Ca+2H_2O\rightarrow Ca\left(OH\right)_2+H_2\uparrow\) (2)
a) Đặt \(\left\{{}\begin{matrix}n_{Na}=a\left(mol\right)\\n_{Ca}=b\left(mol\right)\end{matrix}\right.\) \(\Rightarrow23a+40b=1,6\) (*)
Ta có: \(n_{H_2\left(1\right)}+n_{H_2\left(2\right)}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\)
\(\Rightarrow\dfrac{1}{2}a+b=0,25\) (**)
Từ (*) và (**) \(\Rightarrow\) Hệ có nghiệm âm
*Bạn xem lại đề !!