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\(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\uparrow\)
\(nH_2=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)
\(nH_2SO_4=nH_2=0,15\left(mol\right)\)
\(nH_2SO_{4\left(15\%\right)}=\dfrac{0,15.15}{100}=0,0225\left(mol\right)\)
\(mH_2SO_4=0,0225.98=2,205\left(g\right)\)
\(nAl=\dfrac{2}{3}.0,15=0,1\left(mol\right)\)
\(mAl=0,1.27=2,7\left(g\right)\)
2Al+3H2SO4->Al2(SO4)3+3H2
0,1----------------------0,075----0,15
n H2=0,15 mol
=>mAl=0,1.27=2,7g
=>m Al2(SO4)3=0,075.342=25,65g
a) PTHH: \(2Al+3H_2SO_2\rightarrow Al_2\left(SO_4\right)_3+3H_2\)
b) \(n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)
\(n_{Al}=\dfrac{2}{3}.0,15=0,1\left(mol\right)\)
\(m_{Al}=0,1.27=2,7\left(g\right)\)
c) \(n_{Al_2\left(SO_4\right)_3}=\dfrac{1}{2}.0,1=0,05\left(mol\right)\)
\(m_{Al_2\left(SO_4\right)_3}=0,05.342=17,1\left(g\right)\)
a) Fe + 2HCl --> FeCl2 + H2
b) \(n_{H_2}=\dfrac{1,12}{22,4}=0,05\left(mol\right)\)
PTHH: Fe + 2HCl --> FeCl2 + H2
0,05<-0,1<-----------0,05
=> m = 0,05.56 = 2,8 (g)
c) \(m_{HCl}=0,1.36,5=3,65\left(g\right)\Rightarrow m_{dd.HCl}=\dfrac{3,65.100}{10}=36,5\left(g\right)\)
a. PTHH:
Al+H2SO4-->AlSO4+H2
b.Theo ĐLBTKL, ta có:
mAl+mH2SO4=mAl2SO4+mH2
=>mH2SO4=mAl2SO4+mH2-mAl
=171+3-27=147 (g)
\(n_{Zn}=\dfrac{19.5}{65}=0.3\left(mol\right)\)
\(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)
\(0.3........................0.3..........0.3\)
\(m_{ZnSO_4}=0.3\cdot161=48.3\left(g\right)\)
\(V_{H_2}=0.3\cdot22.4=6.72\left(l\right)\)
\(n_{CuO}=\dfrac{16}{80}=0.2\left(mol\right)\)
\(CuO+H_2\underrightarrow{^{^{t^0}}}Cu+H_2O\)
\(0.2..........0.3\)
\(LTL:\dfrac{0.2}{1}< \dfrac{0.3}{1}\Rightarrow H_2dư\)
\(m_{H_2\left(dư\right)}=\left(0.3-0.2\right)\cdot2=0.2\left(g\right)\)
a) $Zn + H_2SO_4 → ZnSO_4 + H_2$
b) n ZnSO4 = n Zn = 19,5/65 = 0,3(mol)
=> m ZnSO4 = 0,3.161 = 48,3(gam)
c) n H2 = n Zn = 0,3(mol)
V H2 = 0,3.22,4 = 6,72 lít
c)
$CuO + H_2 \xrightarrow{t^o} Cu + H_2O$
n CuO = 16/80 = 0,2(mol) < n H2 = 0,3 nên H2 dư
n H2 pư = n CuO = 0,2(mol)
=> m H2 dư = (0,3 - 0,2).2 = 0,2(gam)
a.b.c.\(n_{Al}=\dfrac{2,7}{27}=0,1mol\)
\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
0,1 0,1 0,15 ( mol )
\(V_{H_2}=0,15.22,4=3,36l\)
\(m_{AlCl_3}=0,1.133,5=13,35g\)
d.\(Fe_2O_3+3H_2\rightarrow\left(t^o\right)2Fe+3H_2O\)
0,15 0,1 ( mol )
\(m_{Fe}=0,1.56=5,6g\)
a.b.\(n_{Al}=\dfrac{m_{Al}}{M_{Al}}=\dfrac{5,4}{27}=0,2mol\)
\(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)
0,2 0,3 ( mol )
\(V_{H_2}=n_{H_2}.22,4=0,3.22,4=6,72l\)
c.\(n_{CuO}=\dfrac{m_{CuO}}{M_{CuO}}=\dfrac{32}{80}=0,4mol\)
\(CuO+H_2\rightarrow Cu+H_2O\)
0,4 < 0,3 ( mol )
0,3 0,3 0,3 ( mol )
\(m_A=m_{CuO\left(du\right)}+m_{Cu}=\left[\left(0,4-0,3\right).80\right]+\left(0,3.64\right)=8+19,2=27,2g\)
a) 2Al + 3H2SO4 --> Al2(SO4)3 + 3H2
b) \(n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)
PTHH: 2Al + 3H2SO4 --> Al2(SO4)3 + 3H2
0,2<----0,3<-----------------0,3
=> mAl = 0,2.27 = 5,4 (g)
c) \(m_{dd.H_2SO_4}=\dfrac{0,3.98}{30\%}=98\left(g\right)\)