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a)
$n_{HCl} = \dfrac{3,65}{36,5} = 0,1(mol)$
$CaCO_3 + 2HCl \to CaCl_2 + CO_2 + H_2O$
$n_{CaCO_3} = n_{CaCl_2} = n_{CO_2} = \dfrac{1}{2}n_{HCl} = 0,05(mol)$
$\%m_{CaCO_3} = \dfrac{0,05.100]{31,1}.100\% = 16,08\%$
$\%m_{Ba(NO_3)_2} = 100\% -16,08\% = 83,92\%$
b)
$m_{dd\ sau\ pư} = 31,1 + 96,1 - 0,05.44 = 125(gam)$
$C\%_{Ba(NO_3)_2} = \dfrac{31,1 - 0,05.100}{125}.100\% = 20,88\%$
$C\%_{CaCl_2} = \dfrac{0,05.111}{125}.100\% = 4,44\%$
\(n_{HCl}=\dfrac{44,8}{22,4}=2\)
\(\Rightarrow m_{HCl}=2.36,5=73g\)
=> \(C\%_{HCl}=\dfrac{73}{73+327}\times100\%=18,25\%\)
b.
\(n_{HCl}=\dfrac{250.18,25\%}{36,5}=1,25mol\)
\(n_{CaCO_3}=\dfrac{50}{100}=0,5mol\)
\(CaCO_3+2HCl\rightarrow CaCl_2+CO_2+H_2O\)
\(n_{CaCl_2}=n_{CO_2}=0,5mol\)
\(n_{HClpu}=0,5.2=1mol\)
\(\Rightarrow n_{HCldu}=1,25-1=0,25\)
\(\Rightarrow m_{ddpu}=50+250-0,5.44=278g\)
\(C\%_{HCl}=\dfrac{0,25.36,5}{278}.100\%=3,28\%\)
\(C\%_{CaCl_2}=\dfrac{0,5.111}{278}.100\%=19,96\%\)
1.
\(m_{HCl}=\dfrac{10,95.75}{100}=8,2125\left(g\right)\)
\(\Rightarrow n_{HCl}=\dfrac{8,2125}{35,5}=0,225\left(mol\right)\)
\(Fe_2O_3+6HCl\rightarrow2FeCl_3+3H_2O\)
\(\Rightarrow n_{Fe_2O_3}=\dfrac{1}{6}n_{HCl}=0,0375\left(mol\right)\)
\(n_{FeCl_3}=\dfrac{1}{3}n_{HCl}=0,075\left(mol\right)\)
\(\Rightarrow C\%=\dfrac{0,075.162,5}{0,0375.160+75}.100\%=15,05\%\)
\(n_{Fe}=\dfrac{19,6}{56}=0,35\left(mol\right)\)
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
0,35--> 0,7-----> 0,35--> 0,35
\(m_{dd.HCl}=\dfrac{0,7.36,5.100\%}{7,3\%}=350\left(g\right)\\ m_{dd}=19,6+350-0,35.2=368,9\left(g\right)\\ C\%_{FeCl_2}=\dfrac{127.0,35.100\%}{368,9}=12,05\%\)
\(m_{HCl}=100.7,3\%=7,3\left(g\right)\Rightarrow n_{HCl}=\dfrac{7,3}{36,5}=0,2\left(mol\right)\)
PT: \(Ba\left(OH\right)_2+2HCl\rightarrow BaCl_2+2H_2O\)
Theo PT: \(n_{BaCl_2}=\dfrac{1}{2}n_{HCl}=0,1\left(mol\right)\)
\(\Rightarrow C\%_{BaCl_2}=\dfrac{0,1.208}{100+100}.100\%=10,4\%\)
a) CaO + 2HCl --> CaCl2 + H2O
CaCO3 + 2HCl --> CaCl2 + CO2 + H2O
b) \(n_{CO_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)
PTHH: CaCO3 + 2HCl --> CaCl2 + CO2 + H2O
_____0,1<-----0,2<-------0,1<-----0,1
=> mCaCO3 = 0,1.100 = 10 (g)
=> mCaO = 15,6 - 10 = 5,6 (g)
b) \(n_{CaO}=\dfrac{5,6}{56}=0,1\left(mol\right)\)
PTHH:CaO + 2HCl --> CaCl2 + H2O
_____0,1--->0,2------>0,1
=> mHCl = (0,2+0,2).36,5 = 14,6 (g)
=> \(m_{ddHCl}=\dfrac{14,6.100}{14,6}=100\left(g\right)\)
mdd sau pư = 15,6 + 100 - 0,1.44 = 111,2 (g)
=> \(C\%\left(CaCl_2\right)=\dfrac{\left(0,1+0,1\right).111}{111,2}.100\%=19,96\%\)
PTHH : CaO + 2HCl ---> CaCl2 + H2O (1)
1 : 2 : 1 : 2
CaCO3 + 2HCl ---> CaCl2 + H2O + CO2 (2)
1 : 2 : 1 : 1 : 1
Ta có \(n_{CO_2}=\dfrac{V}{22.4}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)
=> \(n_{CaCO_3}=0,1\left(mol\right)\)
=> \(m_{CaCO_3}=n.M=0,1.100=10\left(g\right)\)
=> mCaO = 15,6 - 10 = 5,6 (g)
c) \(n_{CaO}=\dfrac{m}{M}=\dfrac{5,6}{56}=0,1\left(mol\right)\)
\(m_{CO_2}=n.M=0,1.44=4,4\left(g\right)\)
Ta có \(m_{HCl}=m_{HCl\left(1\right)}+m_{HCl\left(2\right)}\)
\(=n_{HCl\left(1\right)}.M+n_{HCl\left(2\right)}.M\)
\(0,2.36,5+0,2.36,5=14,6\left(g\right)\)
=> \(m_{ddHCl}=\dfrac{m_{HCl}.100\%}{C\%}=\dfrac{14,6.100\%}{14.6\%}100\left(g\right)\)
\(m_{dd\text{ sau pư}}=m_{ddHCl}+m_{CaO}+m_{CaCO_3}-m_{CO_2}\)
= 100 + 5.6 + 10 - 4,4 = 111.2(g)
=> \(m_{CaCl_2}=m_{CaCl_2\left(1\right)}+m_{CaCl_2\left(2\right)}\)
\(=n_{CaCl_2\left(1\right)}.M+n_{CaCl_2\left(2\right)}.M\)
= 0,1.91 + 0,1.91 = 18,2 (g)
=> \(C\%=\dfrac{m_{CaCl_2}}{m_{\text{dd sau pư}}}.100\%=\dfrac{18,2}{111,2}.100\%=16,37\%\)
\(CaCO_3+ HCl → CaCl_2+H_2O +CO_2\)
\(n_{CaCO_3}=\dfrac{10}{40+12+16.3}=0,1(mol)\)
\(n_{HCl}=\dfrac{146}{1+35,5}=4(mol)\)
\(\Rightarrow n_{HCl_{dư}}=4-0,1=3,9(mol) ; n_{CaCl_2}=0,1(mol)\\\Rightarrow m_{\text{chất tan}} = m_{HCl_{dư}}+m_{CaCl_2}\\=0,39.(35,5+1)+0,1(40+35,5.2)=25,335(g)\)
Vậy...
- Ngu Hóa, sai thì thôi nhé. :<<