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\(n_{CO_2}=\dfrac{2,8}{22,4}=0,125\left(mol\right)\)
PTHH :
\(CO_2+2KOH\rightarrow K_2CO_3+H_2O\)
0,125 0,25 0,125
\(a,C_{M\left(KOH\right)}=\dfrac{0,25}{0,1}=2,5\left(M\right)\)
\(b,C_{M\left(K_2CO_3\right)}=\dfrac{0,125}{0,1}=1,25\left(M\right)\)
1.PTHH :
\(CaCO_3--to->CaO+CO_2\)
\(m_{CaCO_3}=\frac{10^6.80}{100}=0,8.10^6\left(g\right)\)
\(CaCO_3--to->CaO+CO_2\)
100 g____________ 56 g
0,8.\(10^6\) ____________ x
=> x=\(\frac{0,8.10^6.56}{100}=0,448.10^6\left(g\right)\)
=> \(m_{CaO}thuđc=0,448.10^6.90\%=0,4032.10^6\left(g\right)\)
2.
Đặt HX là d2 \(h^2\) HCl và HNO3
=> \(n_{HX}=n_{HNO_3}+n_{HCl}=0,2V+0,2V=0,4V\left(mol\right)\)
\(n_{CaO}=\frac{11,2}{56}=0,2\left(mol\right)\)
\(CaO+2HX-->CaX_2+H_2O\)
0,2_____0,4
=> a) \(V_{HX}=\frac{0,4}{0,4}=1\left(l\right)\)
b) \(n_{Ca^{2+}}=n_{CaO}=0,2\left(mol\right)\)
\(n_{Cl^-}=n_{HCl}=0,2\left(mol\right)\)
\(n_{NO_3^-}=n_{HNO_3}=0,2\left(mol\right)\)
=> \(m_{CaX_2}=m_{Ca^{2+}}+m_{Cl^-}+m_{NO_3^-}=0,2.40+0,2.35,5+0,2.62=27,5\left(g\right)\)
PTHH: \(CH_3COOH+KHCO_3\rightarrow CH_3COOK+H_2O+CO_2\uparrow\)
a) Ta có: \(n_{CH_3COOH}=\dfrac{200\cdot24\%}{60}=0,8\left(mol\right)=n_{KHCO_3}\)
\(\Rightarrow m_{ddKHCO_3}=\dfrac{0,8\cdot100}{16,8\%}\approx476.2\left(g\right)\)
b) Theo PTHH: \(n_{CH_3COOK}=0,8\left(mol\right)=n_{CO_2}\)
\(\Rightarrow\left\{{}\begin{matrix}m_{CH_3COOK}=0,8\cdot98=78,4\left(g\right)\\m_{CO_2}=0,8\cdot44=35,2\left(g\right)\end{matrix}\right.\)
Mặt khác: \(m_{dd}=m_{ddCH_3COOH}+m_{ddKHCO_3}-m_{CO_2}=641\left(g\right)\)
\(\Rightarrow C\%_{CH_3COOK}=\dfrac{78,4}{641}\cdot100\%\approx12,23\%\)
Câu 1:
\(n_{CO_2}=\dfrac{11,2}{22,4}=0,5(mol)\\ a,PTHH:CO_2+Ba(OH)_2\to BaCO_3\downarrow+H_2O\\ \Rightarrow n_{Ba(OH)_2}=n_{BaCO_3}=n_{CO_2}=0,5(mol)\\ \Rightarrow C_{M_{Ba(OH)_2}}=\dfrac{0,5}{0,2}=2,5M\\ m_{BaCO_3}=0,5.197=98,5(g)\\ b,PTHH:Ba(OH)_2+2HCl\to BaCL_2+2H_2O\\ \Rightarrow n_{HCl}=2n_{Ba(OH)_2}=1(mol)\\ \Rightarrow m_{CT_{HCl}}=1.36,5=36,5(g)\\ \Rightarrow m_{dd_{HCl}}=\dfrac{36,5}{20\%}=182,5(g)\)
Câu 2:
\(n_{Fe_2O_3}=\dfrac{32}{160}=0,2(mol)\\ m_{HCl}=\dfrac{292.20\%}{100\%}=58,4(g)\\ \Rightarrow n_{HCl}=\dfrac{58,4}{36,5}=1,6(mol)\\ PTHH:Fe_2O_3+6HCl\to 2FeCl_3+3H_2O\)
Vì \(\dfrac{n_{HCl}}{6}>\dfrac{n_{Fe_2O_3}}{1}\) nên \(HCl\) dư
\(\Rightarrow n_{FeCl_3}=2n_{Fe_2O_3}=0,4(mol);n_{H_2O}=3n_{Fe_3O_3}=0,6(mol)\\ \Rightarrow \begin{cases} m_{CT_{FeCl_3}}=0,4.162,5=65(g)\\ m_{H_2O}=0,6.18=10,8(g) \end{cases}\\ \Rightarrow m_{dd_{FeCl_3}}=32+292-10,8=313,2(g)\\ \Rightarrow C\%_{FeCl_3}=\dfrac{65}{313,2}.100\%\approx20,75\%\)
nNa2O = 0.1 mol
Na2O + H2O --> 2NaOH
0.1_____________0.2
2NaOH + CO2 --> Na2CO3 + H2O
0.2______0.1_______0.1
VCO2 = 2.24 l
mCO2 = 4.4 g
mdd sau phản ứng = 4.4 + 200 = 204.4 g
mNa2CO3 = 10.6 g
C%Na2CO3 = 10.6/204.4*100% = 5.18%