Cho 3,375g al vào bình đựng 300g dung dịch H2SO4 4,9% đến khi phản ứng xảy ra hoàn toàn. Tính nồng độ phần trăm của dung dịch sau phản ứng
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\(m_{CH_3COOH}=6\%.200=12\left(g\right)\\ \rightarrow n_{CH_3COOH}=\dfrac{12}{60}=0,2\left(mol\right)\\ n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\)
PTHH: \(2CH_3COOH+Zn\rightarrow\left(CH_3COO\right)_2Zn+H_2\uparrow\)
LTL: \(\dfrac{0,2}{2}>0,2\rightarrow\) Zn dư
Theo pthh: \(n_{\left(CH_3COO\right)_2Zn}=n_{Zn\left(pư\right)}=n_{H_2}=\dfrac{1}{2}n_{CH_3COOH}=\dfrac{1}{2}.0,2=0,1\left(mol\right)\)
\(\rightarrow\left\{{}\begin{matrix}m_{H_2}=0,1.2=0,2\left(g\right)\\m_{Zn\left(pư\right)}=0,1.65=6,5\left(g\right)\\m_{\left(CH_3COO\right)_2Zn}=0,1.183=18,3\left(g\right)\end{matrix}\right.\)
\(\rightarrow m_{dd}=200+6,5-0,2=206,3\left(g\right)\\ \rightarrow C\%_{\left(CH_3COO\right)_2Zn}=\dfrac{18,3}{206,3}=8,87\%\)
a)
$Mg + H_2SO_4 \to MgSO_4 + H-2$
b) $n_{H_2SO_4} = n_{Mg} = \dfrac{4,8}{24} = 0,2(mol)$
$C\%_{H_2SO_4} = \dfrac{0,2.98}{200}.100\% = 9,8\%$
$n_{H_2} = n_{Mg} = 0,2(mol)$
$\Rightarrow m_{dd\ A} = 4,8 + 200 - 0,2.2 = 204,4(gam)$
$C\%_{MgSO_4} = \dfrac{0,2.120}{204,4}.100\% = 11,7\%$
c) $V_{H_2} = 0,2.22,4 = 4,48(lít)$
nCaCO3=8,4:(40+12+16.3)=0,084 mol
nH2SO4= 0,5.1=0,5 mol
PTHH: H2SO4+ CaCO3 --> CaSO4↓ + CO2 +H2O
theo đề: 0,5 mol: 0,084 mol
=> H2SO4 de theo CaCO3
phản ứng : 0,084mol<----0,084 mol---> 0,084mol
=> CM=\(\frac{0,084}{0,5}=0,168M\)
nNa2CO3 = 10,6 / 106 = 0,1 (mol)
Na2CO3 + 2CH3COOH -> 2CH3COONa + H2O + CO2
0,1 0,2 0,2 0,1
mdd CH3COOH = 0,2 * 60 / 5 * 100 = 240 (gam)
CO2 + Ca(OH)2 -> CaCO3 + H2O
0,1 0,1
mCaCO3 = 0,1 * 100 = 10 (gam)
mdd = 240 + 10,6 - 0,1 * 44 = 246,2 (gam)
C% = 82 * 0,2 / 246,2 * 100% = 6,66%
\(n_{Na_2CO_3}=\dfrac{10,6}{106}=0,1\left(mol\right)\)
PTHH:
Na2CO3 + 2CH3COOH ---> 2CH3COONa + CO2 + H2O
0,1---------->0,2----------------->0,2--------------->0,1
CO2 + Ca(OH)2 ---> CaCO3 + H2O
0,1------------------------->0,1
=> \(\left\{{}\begin{matrix}m_{ddCH_3COOH}=\dfrac{0,2.60}{5\%}=240\left(g\right)\\m_{CaCO_3}=0,1.100=10\left(g\right)\end{matrix}\right.\)
\(m_{dd}=10,6+240-0,1.44=246,2\left(g\right)\\ C\%_{CH_3COONa}=\dfrac{82.0,2}{246,2}.100\%=6,66\%\)
\(n_{HCl}=\dfrac{44,8}{22,4}=2\)
\(\Rightarrow m_{HCl}=2.36,5=73g\)
=> \(C\%_{HCl}=\dfrac{73}{73+327}\times100\%=18,25\%\)
b.
\(n_{HCl}=\dfrac{250.18,25\%}{36,5}=1,25mol\)
\(n_{CaCO_3}=\dfrac{50}{100}=0,5mol\)
\(CaCO_3+2HCl\rightarrow CaCl_2+CO_2+H_2O\)
\(n_{CaCl_2}=n_{CO_2}=0,5mol\)
\(n_{HClpu}=0,5.2=1mol\)
\(\Rightarrow n_{HCldu}=1,25-1=0,25\)
\(\Rightarrow m_{ddpu}=50+250-0,5.44=278g\)
\(C\%_{HCl}=\dfrac{0,25.36,5}{278}.100\%=3,28\%\)
\(C\%_{CaCl_2}=\dfrac{0,5.111}{278}.100\%=19,96\%\)
\(m_{NaOH\left(A\right)}=20.5\%=1\left(g\right)\)
Trong B:
gọi x là khối lượng Na2O thêm vào , x>0 (g)
\(10\%=\dfrac{\dfrac{80}{62}x+1}{x+20}\)
\(\rightarrow x=0,84\left(g\right)\)
Vậy khối Na2O thêm vào dd A là 0,84 (g)
b, \(m_{KOH\left(A\right)}=2\%.20=0,4\left(g\right)\)
\(C\%_{KOH\left(B\right)}=\dfrac{0,4}{20+0,84}.100\%=1,92\%\)
\(n_K=\dfrac{5,85}{39}=0,15\left(mol\right)\)
PTHH: 2K + 2H2O --> 2KOH + H2
_____0,15------------->0,15-->0,075
=> VH2 = 0,075.22,4 =1,68(l)
mdd = 5,85 + 100 - 0,075.2 = 105,7(g)
=> \(C\%=\dfrac{0,15.56}{105,7}.100\%=7,95\%\)
PTHH: \(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\uparrow\)
Ta có: \(\left\{{}\begin{matrix}n_{Al}=\dfrac{3,375}{27}=0,125\left(mol\right)\\n_{H_2SO_4}=\dfrac{300\cdot4,9\%}{98}=0,15\left(mol\right)\end{matrix}\right.\)
Xét tỉ lệ: \(\dfrac{0,125}{2}>\dfrac{0,15}{3}\) \(\Rightarrow\) Al còn dư, H2SO4 p/ứ hết
\(\Rightarrow\left\{{}\begin{matrix}n_{H_2}=0,15\left(mol\right)\\n_{Al\left(dư\right)}=0,025\left(mol\right)\\n_{Al_2\left(SO_4\right)_3}=0,05\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{H_2}=0,15\cdot2=0,3\left(g\right)\\m_{Al\left(dư\right)}=0,025\cdot27=0,672\left(g\right)\\m_{Al_2\left(SO_4\right)_3}=0,05\cdot342=17,1\left(g\right)\end{matrix}\right.\)
Mặt khác: \(m_{dd}=m_{Al}+m_{ddH_2SO_4}-m_{Al\left(dư\right)}-m_{H_2}=302,403\left(g\right)\)
\(\Rightarrow C\%_{Al_2\left(SO_4\right)_3}=\dfrac{17,1}{302,403}\cdot100\%\approx5,65\%\)