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Bài 3 :
\(n_{Al}=\dfrac{2,7}{27}=0,1\left(mol\right)\)
Pt : \(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2|\)
2 3 1 3
0,1 0,15 0,05 0,15
a) \(n_{H2}=\dfrac{0,1.3}{2}=0,15\left(mol\right)\)
\(m_{H2}=0,15.2=0,3\left(g\right)\)
\(V_{H2\left(dktc\right)}=0,15.22,4=3,36\left(l\right)\)
b) \(n_{H2SO4}=\dfrac{0,1.3}{2}=0,15\left(mol\right)\)
⇒ \(m=0,15.98=14,7\left(g\right)\)
\(C_{ddH2SO4}=\dfrac{14,7.100}{200}=7,35\)0/0
c) \(n_{Al2\left(SO4\right)3}=\dfrac{0,15.1}{3}=0,05\left(mol\right)\)
⇒ \(m_{Al2\left(SO4\right)3}=0,05.342=17,1\left(g\right)\)
\(m_{ddspu}=2,7+200-0,3=302,4\left(g\right)\)
\(C_{Al2\left(SO4\right)3}=\dfrac{17,1.100}{302,4}=5,65\)0/0
Chúc bạn học tốt
Mình xin lỗi bạn nhé , bạn sửa lại giúp mình :
\(m_{ddspu}=2,7+200-0,3=202,4\left(g\right)\)
\(C_{Al2\left(SO4\right)3}=\dfrac{17,1.100}{202,4}=8,45\)0/0
a,\(n_{Al}=\dfrac{2,7}{27}=0,1\left(mol\right)\)
PTHH: 2Al + 3H2SO4 → Al2(SO4)3 + 3H2
Mol: 0,1 0,15 0,05 0,15
\(m_{H_2}=0,15.2=0,3\left(g\right)\)
\(V_{H_2}=0,15.24,79=3,7185\left(l\right)\)
b, \(m_{ddH_2SO_4}=\dfrac{0,15.98.100\%}{200}=7,35\%\)
c, mdd sau pứ = 2,7 + 200 - 0,3 = 202,4 (g)
\(C\%_{ddAl_2\left(SO_4\right)_3}=\dfrac{0,05.342.100\%}{202,4}=8,45\%\)
a) 2Al + 3H2SO4 --> Al2(SO4)3 + 3H2
b) \(n_{H_2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\)
PTHH: 2Al + 3H2SO4 --> Al2(SO4)3 + 3H2
____\(\dfrac{4}{15}\)<----0,4<--------------------0,4
=> \(m_{Al}=\dfrac{4}{15}.27=7,2\left(g\right)\)
c) \(C_{M\left(H_2SO_4\right)}=\dfrac{0,4}{0,15}=2,667M\)
\(a,n_{H_2SO_4}=0,5\cdot0,1=0,05\left(mol\right)\\ PTHH:2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\\ \Rightarrow n_{H_2}=n_{H_2SO_4}=0,05\left(mol\right)\\ \Rightarrow V_{H_2\left(đktc\right)}=0,05\cdot22,4=1,12\left(l\right)\\ b,n_{Al}=\dfrac{2}{3}n_{H_2SO_4}=\dfrac{1}{30}\left(mol\right)\\ \Rightarrow m_{Al}=\dfrac{1}{30}\cdot27=0,9\left(g\right)\\ c,n_{Al_2\left(SO_4\right)_3}=\dfrac{1}{3}n_{H_2SO_4}\approx0,017\left(mol\right)\\ \Rightarrow C_{M_{Al_2\left(SO_4\right)_3}}=\dfrac{0,017}{0,1}\approx0,17M\)
a) \(n_{Na_2CO_3}=\dfrac{21,2}{106}=0,2\left(mol\right)\)
PTHH: Na2CO3 + 2HCl --> 2NaCl + CO2 + H2O
0,2------------------>0,4---->0,2
mdd sau pư = 200 + 21,2 - 0,2.44 = 212,4(g)
=> \(C\%\left(NaCl\right)=\dfrac{0,4.58,5}{212,4}.100\%=11,017\%\)
$PTHH:Na_2CO_3+2HCl\to 2NaCl+H_2O+CO_2\uparrow$
$n_{Na_2CO_3}=\dfrac{21,2}{106}=0,2(mol)$
Theo PT: $n_{NaCl}=n_{CO_2}=0,2(mol)$
$\Rightarrow m_{NaCl}=0,4.58,5=23,4(g);m_{CO_2}=0,2.44=8,8(g)$
$\Rightarrow C\%_{NaCl}=\dfrac{23,4}{21,2+200-8,8}.100\%\approx 11,01\%$
a)
$Zn + H_2SO_4 \to ZnSO_4 + H_2$
Theo PTHH :
$n_{H_2} = n_{Zn} = \dfrac{6,5}{65} = 0,1(mol)$
$V_{H_2} = 0,1.22,4 = 2,24(lít)$
b) $n_{H_2SO_4} = n_{Zn} = 0,1(mol)$
$V_{dd\ H_2SO_4} = \dfrac{0,1}{1} = 0,1(lít)$
c) $n_{ZnSO_4} = 0,1(mol) \Rightarrow m_{ZnSO_4} = 0,1.161 = 16,1(gam)$
d) $C_{M_{ZnSO_4}} = \dfrac{0,1}{0,1} = 1M$
\(n_{CuO}=\dfrac{1.6}{80}=0.02\left(mol\right)\)
\(CuO+H_2SO_4\rightarrow CuSO_4+H_2O\)
\(0.02.......0.02.................0.02\)
\(m_{H_2SO_4}=0.02\cdot98=1.96\left(g\right)\)
\(m_{dd_{H_2SO_4}}=\dfrac{1.96}{20\%}=9.8\left(g\right)\)
\(m_{\text{dung dịch sau phản ứng }}=1.6+9.8=11.4\left(g\right)\)
\(C\%_{CuSO_4}=\dfrac{0.02\cdot160}{11.4}=28.07\%\)
\(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)
a) \(n_{Al}=\dfrac{10,8}{27}=0,4\left(mol\right)\)
Theo pt: \(n_{H_2}=\dfrac{3}{2}n_{Al}=\dfrac{3}{2}.0,4=0,6\left(mol\right)\)
\(V_{H_2}=0,6.22,4=13,44\left(l\right)\)
b) Theo pt: \(n_{H_2SO_4}=\dfrac{3}{2}n_{Al}=\dfrac{3}{2}.0,4=0,6\left(mol\right)\)
\(m_{H_2SO_4}=0,6.98=58,8g\)
\(C_{\%}dd_{H_2SO_4}=\dfrac{m_{ct}}{m_{dd}}.100\%=\dfrac{58,8}{200}.100\%=29,4\%\)
c) Theo pt: \(n_{Al_2\left(SO_4\right)_3}=\dfrac{n_{Al}}{2}=0,2\left(mol\right)\)
\(m_{Al_2\left(SO_4\right)_3}=0,2.342=68,4\left(g\right)\)
Áp dụng định luật bảo toàn khối lượng
\(m_{dd_{Al_2\left(SO_4\right) _3}}=m_{Al}+m_{dd_{H_2SO_4}}-m_{H_2}\)
\(=10,8+200-0,6.2=209,6g\)
\(C_{\%_{Al_2\left(SO_4\right)_3}}=\dfrac{68,4}{209,6}.100\%\approx32,6\%\)