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a)\(2Na+2H2O---.2NaOH+H2\)
b)\(n_{Na}=\frac{4,6}{23}=0,2\left(mol\right)\)
\(n_{NaOH}=n_{Na}=0,2\left(mol\right)\)
\(m_{NaOH}=0,2.40=8\left(g\right)\)
\(n_{H2}=\frac{1}{2}n_{Na}=0,1\left(mol\right)\)
\(V_{H2}=0,1.22,4=2,24\left(l\right)\)
c)\(n_{Na}=\frac{2,3}{23}=0,1\left(mol\right)\)
\(n_{H2O}=\frac{3,6}{18}=0,2\left(mol\right)\)
Tỉ lệ Na:H2O=1:1
\(n_{Na}< n_{H2O}\)
=>H2O dư
\(n_{H2O}=n_{Na}=0,1\left(mol\right)\)
\(n_{H2O}dư=0,2-0,1=0,1\left(mol\right)\)
\(m_{H2O}=0,1.18=1,8\left(g\right)\)
$a)n_{Fe}=\dfrac{42}{56}=0,75(mol)$
$Fe_2O_3+3H_2\xrightarrow{t^o}2Fe+3H_2O$
$\Rightarrow n_{Fe_2O_3}=0,5n_{Fe}=0,375(mol)$
$\Rightarrow m_{Fe_2O_3}=0,375.160=60(g)$
$b)n_{H_2O}=1,5n_{Fe}=1,125(mol)$
$\Rightarrow m_{H_2O}=1,125.18=20,25(g)$
a) nZn= 13/65 = 0,2 (mol)
Zn + 2HCl → ZnCl2 +H2
0,2 →0,4
mHCl= 0,4. 36,5=14,6(gam)
b) nH2=0,4/2 =0,2 (mol)
mZn= 0,2 .65=13 (g)
mHCl = 0,4.36,5 =14,6(gam)
1. Na + 1/2O2 -> NaO
Al2O3 + 6HCl -> 2AlCl3 + 3H2O
AgNO3 + NaCl -> AgCl + NaNO3
CuSO4 + 2NaOH -> Na2SO4 + Cu(OH)2
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, \(n_{NaOH}=n_{Na}=0,1\left(mol\right)\Rightarrow m_{NaOH}=0,1.40=4\left(g\right)\)
\(2Na+2H_2O\rightarrow2NaOH+H_2\\ n_{Na}=\dfrac{4,6}{23}=0,2\left(mol\right)\\ n_{H_2O}=\dfrac{1,8}{18}=0,1\left(mol\right)\\ LTL:\dfrac{0,2}{2}>\dfrac{0,1}{2}\\ \Rightarrow Nadư\\ n_{Na\left(pứ\right)}=n_{H_2O}=0,1\left(mol\right)\\ n_{Na\left(dư\right)}=0,2-0,1=0,1\left(mol\right)\\ \Rightarrow m_{Na}=0,1.23=2,3\left(g\right)\\ n_{H_2}=\dfrac{1}{2}n_{H_2O}=0,05\left(mol\right)\\ \Rightarrow V_{H_2}=0,05.22,4=1,12\left(l\right)\)
a)
$Fe + 2HCl \to FeCl_2 + H_2$
$n_{Fe} = n_{H_2} = \dfrac{3,36}{22,4} = 0,15(mol)$
$m_{Fe} = 0,15.56 = 8,4(gam)$
b) $n_{HCl} = 2n_{H_2} = 0,3(mol)$
$\Rightarrow m_{HCl} = 0,3.36,5 = 10,95(gam)$
c)
Cách 1 : $n_{FeCl_2} = n_{H_2} = 0,15(mol) \Rightarrow m_{FeCl_2} = 0,15.127 = 19,05(gam)$
Cách 2 : Bảo toàn khối lượng, $m_{FeCl_2} = 8,4 + 10,95 - 0,15.2 = 19,05(gam)$
\(n_{Al}=\dfrac{2,7}{27}=0,1\left(mol\right)\)
PTHH: 2Al + 6HCl --> 2AlCl3 + 3H2
_____0,1--->0,3
=> mHCl = 0,3.36,5 = 10,95(g)
nNa=4,6/23=0,2(mol)
PT:2Na+2H2O->2NaOH+H2
Theo Pt: nH2= 1/2n Na =1/2.0,2=0,1 (mol)
a,=>VH2=0,1.22,4=2,24(l)
b,nH2=3,36/22,4=0,15 (mol)
pt:2H2+O2->2H2O
Theo Pt: n H2O =nH2 =0,15
b,->m H2O =0,15.18=2,7 (g)
c, Tương tự nhé =)))