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\(2Na+2H_2O\rightarrow2NaOH+H_2\\ Na_2O+H_2O\rightarrow2NaOH\\ n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\\ n_{Na}=2.0,3=0,6\left(mol\right)\\ a,m_{Na}=0,6.23=13,8\left(g\right)\\ m_{Na_2O}=26,2-13,8=12,4\left(g\right)\\b, n_{Na_2O}=\dfrac{12,4}{62}=0,2\left(mol\right)\\ n_{NaOH\left(tổng\right)}=n_{Na}+2.n_{Na_2O}=0,6+\dfrac{12,4}{62}=0,8\left(mol\right)\\ m_{c.tan}=m_{NaOH}=0,8.40=32\left(g\right)\\ c,m_{ddNaOH}=m_{hh}+m_{H_2O}-m_{H_2}=26,2+200-0,3.2=225,6\left(g\right)\\ C\%_{ddNaOH}=\dfrac{32}{225,6}.100\approx14,185\%\)
a, PT: \(Zn+2HCl\rightarrow ZnCl_2+H_2\)
\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
Gọi: \(\left\{{}\begin{matrix}n_{Zn}=x\left(mol\right)\\n_{Al}=y\left(mol\right)\end{matrix}\right.\) ⇒ 65x + 27y = 17,05 (1)
Ta có: \(n_{H_2}=\dfrac{9,52}{22,4}=0,425\left(mol\right)\)
Theo PT: \(n_{H_2}=n_{Zn}+\dfrac{3}{2}n_{Al}=x+\dfrac{3}{2}y=0,425\left(2\right)\)
Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}n_{Zn}=0,2\left(mol\right)\\n_{Al}=0,15\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}m_{Zn}=0,2.65=13\left(g\right)\\m_{Al}=0,15.27=4,05\left(g\right)\end{matrix}\right.\)
b, Theo PT: \(\left\{{}\begin{matrix}n_{ZnCl_2}=n_{Zn}=0,2\left(mol\right)\\n_{AlCl_3}=n_{Al}=0,15\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}C_{M_{ZnCl_2}}=\dfrac{0,2}{0,5}=0,4\left(M\right)\\C_{M_{AlCl_3}}=\dfrac{0,15}{0,5}=0,3\left(M\right)\end{matrix}\right.\)
c, Ta có: m dd HCl = 1,05.500 = 525 (g)
m dd sau pư = mhh + m dd HCl - mH2 = 541,2 (g)
\(\Rightarrow\left\{{}\begin{matrix}C\%_{ZnCl_2}=\dfrac{0,2.136}{541,2}.100\%\approx5,03\%\\C\%_{AlCl_3}=\dfrac{0,15.133,5}{541,2}.100\%\approx3,7\%\end{matrix}\right.\)
1 ) CAO +H2O => CA(OH)2 (1)
2K + 2H2O => 2KOH + H2(2)
n (H2) =1,12/22,4 =0,05
theo ptpư 2 : n(K) = 2n (h2) =2.0.05=0,1(mol)
=> m (K) =39.0,1=3,9 (g)
% K= 3,9/9,5 .100% =41,05%
%ca =100%-41,05%=58,95%
xo + 2hcl =>xcl2 +h2o
10,4/X+16 15,9/x+71
=> giải ra tìm đc X bằng bao nhiêu thì ra
\(a) n_{Mg} = a(mol) ; n_{Fe} = b(mol)\\ \Rightarrow 24a + 56b = 9,6(1)\\ 2Mg + O_2 \xrightarrow{t^o} 2MgO\\ 3Fe + 2O_2 \xrightarrow{t^o} Fe_3O_4\\ n_{O_2} = 0,5a + \dfrac{2}{3}b = \dfrac{2,8}{22,4} = 0,125(2)\\ (1)(2)\Rightarrow a = 0,05 ; b = 0,15\\ m_{Mg} = 0,05.24 = 1,2(gam) ; m_{Fe} = 0,15.56 = 8,4(gam)\\ b) m_{oxit} = m_A + m_{O_2} = 9,6 + 0,125.32 = 13,6(gam)\)
a) PTHH : \(FeO+H_2-t^o->Fe+H_2O\)
\(CuO+H_2-t^o->Cu+H_2O\)
Đặt \(\hept{\begin{cases}n_{FeO}=x\left(mol\right)\\n_{CuO}=y\left(mol\right)\end{cases}}\) => \(72x+80y=11,2\left(I\right)\)
Có : \(m_{O\left(lấy.đi\right)}=m_{giảm}=1,92\left(g\right)\)
=> \(n_{O\left(lấy.đi\right)}=\frac{1,92}{16}=0,12\left(mol\right)\) Vì H% = 80% => Thực tế : \(n_{O\left(hh\right)}=\frac{0,12}{80}\cdot100=0,15\left(mol\right)\)
BT Oxi : \(x+y=0,15\left(II\right)\)
Từ (I) và (II) suy ra : \(\hept{\begin{cases}x=0,1\\y=0,05\end{cases}}\)
=> \(\hept{\begin{cases}m_{FeO}=7,2\left(g\right)\\m_{CuO}=4\left(g\right)\end{cases}}\)
b) PTHH : \(Fe+H_2SO_4-->FeSO_4+H_2\)
BT Fe : \(n_{Fe}=n_{FeO}=0,1\left(mol\right)\)
Theo pthh : \(n_{H_2}=n_{Fe}=0,1\left(mol\right)\)
=> \(V_{H_2}=2,24\left(l\right)\)
BT Cu : \(n_{Cu}=n_{CuO}=0,05\left(mol\right)\)
=> \(m_{CR\left(ko.tan\right)}=0,05\cdot64=3,2\left(g\right)\)
1.
2Na + 2H2O\(\rightarrow\)2NaOH + H2 (1)
Na2O + H2O \(\rightarrow\)2NaOH (2)
nH2=\(\dfrac{3,36}{22,4}=0,15\left(mol\right)\)
Theo PTHH 1 ta có:
2nH2=nNa=0,3(mol)
mNa=0,3.23=6,9(g)
mNa2O=13,1-6,9=6,2(g)
nNa2O=\(\dfrac{6,2}{62}=0,1\left(mol\right)\)
Theo PTHH 1 và 2 ta có:
nNa=nNaOH=0,3(mol)
2nNa2O=nNaOH=0,2(mol)
mNaOH=(0,2+0,3).40=20(g)
C% dd NaOH=\(\dfrac{20}{13,1+100-0,15.2}.100\%=17,73\%\)
2, gọi x la so mol cua Al
y la so mol cua Fe
2Al + 3H2SO4 \(\rightarrow\) Al2(SO4)3 + 3H2
de: x \(\rightarrow\) 1,5x \(\rightarrow\) 1,5x
Fe + H2SO4 \(\rightarrow\) FeSO4 + H2
de: y \(\rightarrow\) y \(\rightarrow\) y
\(n_{H_2}=\dfrac{13,44}{22,4}=0,6\left(mol\right)\)
Ta co: 27x + 56y = 22,2
1,5x + y = 0,6
=> x= 0,2 y= 0,3
b, \(\%m_{Al}=\dfrac{27.0,2}{22,2}.100\%\approx24,32\%\)
\(\%m_{Fe}=100-24,32\approx75,68\%\)
c, \(m_{H_2SO_4}=98.\left(0,2.1,5+0,3\right)=58,8g\)
\(C\%_{H_2SO_4}=\dfrac{58,8}{100}.100\%=58,8\%\)
$a\big)$
$M_A=9,4.2=18,8(g/mol)$
$\to \dfrac{n_{CO_2}}{n_{H_2}}=\dfrac{18,8-2}{44-18,8}=\dfrac{2}{3}$
Mà $n_{CO_2}+n_{H_2}=\dfrac{11,2}{22,4}=0,5(mol)$
\(\begin{array} {l} \to n_{CO_2}=0,2(mol);n_{H_2}=0,3(mol)\\ Fe+2HCl\to FeCl_2+H_2\\ FeCO_3+2HCl\to FeCl_2+CO_2+H_2O\\ \text{Theo PT: }n_{Fe}=n_{H_2}=0,3(mol);n_{FeCO_3}=n_{CO_2}=0,2(mol)\\ \to m=0,3.56+0,2.116=40(g) \end{array}\)
$b\big)$
Đổi $400ml=0,4l$
\(\begin{array} {l} \text{Theo PT: }n_{FeCl_2}=n_{H_2}+n_{CO_2}=0,5(mol)\\ \to C_{M\,FeCl_2}=\dfrac{0,5}{0,4}=1,25M \end{array}\)
$c\big)$
\(\begin{array}{l} m_{dd\,FeCl_2}=\dfrac{400}{1,2}\approx 333,33(g)\\ \to C\%_{FeCl_2}=\dfrac{0,5.127}{333,33}.100\%=19,05\%\end{array}\)
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