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\(a)Ba\left(OH\right)_2+Na_2CO_3\rightarrow2NaOH+BaCO_3\\ n_{Ba\left(OH\right)_2}=0,2.2=0,4mol\\ n_{BaCO_3}=n_{Na_2CO_3}=n_{Ba\left(OH\right)_2}=0,4mol\\ m_{BaCO_3}=0,4.171=68,4g\\ b)V_{Na_2CO_3}=\dfrac{0,4}{1}=0,4l\\ c)n_{NaOH}=2n_{Ba\left(OH\right)_2}=0,8mol\\ C_{M\left(NaOH\right)}=\dfrac{0,8}{0,2+0,4}=\dfrac{4}{3}M\)
\(n_{CO_2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\)
PT: \(CO_2+Ca\left(OH\right)_2\rightarrow CaCO_3+H_2O\)
Theo PT: \(n_{Ca\left(OH\right)_2}=n_{CaCO_3}=n_{CO_2}=0,25\left(mol\right)\)
a, \(C_{M_{Ca\left(OH\right)_2}}=\dfrac{0,25}{0,1}=2,5\left(M\right)\)
b, \(m_{CaCO_3}=0,25.100=25\left(g\right)\)
c, \(Ca\left(OH\right)_2+2HCl\rightarrow CaCl_2+2H_2O\)
Theo PT: \(n_{HCl}=2n_{Ca\left(OH\right)_2}=0,5\left(mol\right)\)
\(\Rightarrow m_{ddHCl}=\dfrac{0,5.36,5}{20\%}=91,25\left(g\right)\)
\(n_{CuO}=\dfrac{8}{80}=0,1\left(mol\right)\)
a) Pt : \(CuO+2HCl\rightarrow CuCl_2+H_2O|\)
1 2 1 1
0,1 0,2 0,1
b) \(n_{HCl}=\dfrac{0,1.2}{1}=0,2\left(mol\right)\)
⇒ \(m_{HCl}=0,2.36,5=7,3\left(g\right)\)
\(C_{ddHCl}=\dfrac{7,3.100}{200}=3,65\)0/0
c) \(n_{CuCl2}=\dfrac{0,2.1}{2}=0,1\left(mol\right)\)
⇒ \(m_{CuCl2}=0,1.135=13,5\left(g\right)\)
Chúc bạn học tốt
\(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)
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
\(m_{dd_{Al_2\left(SO_4\right) _3}}=m_{Al}+m_{dd_{H_2SO_4}}-m_{H_2}\)
\(=10,8+200-0,6.2=209,6g\)
\(C_{\%_{Al_2\left(SO_4\right)_3}}=\dfrac{68,4}{209,6}.100\%\approx32,6\%\)
\(m_{CH_3COOH}=150.12\%=18g\)
\(n_{CH_3COOH}=\dfrac{18}{60}=0,3mol\)
\(2CH_3COOH+Na_2CO_3\rightarrow2CH_3COONa+CO_2+H_2O\)
0,3 0,15 0,3 0,15 ( mol )
\(m_{ddNa_2CO_3}=\left(0,15.106\right):10,6\%=150g\)
\(V_{CO_2}=0,15.22,4=3,36l\)
\(m_{CH_3COONa}=0,3.82=24,6g\)
\(m_{ddspứ}=150+150-0,15.44=293,4g\)
\(C\%_{CH_3COONa}=\dfrac{24,6}{293,4}.100=8,28\%\)
a) PTHH: \(Fe+2HCl\rightarrow FeCl_2+H_2\)
b) Ta có: \(\left\{{}\begin{matrix}n_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right)\\n_{HCl}=\dfrac{200\cdot10\%}{36,5}=\dfrac{40}{73}\left(mol\right)\end{matrix}\right.\)
Xét tỉ lệ: \(\dfrac{0,2}{1}< \dfrac{\dfrac{40}{73}}{2}\) \(\Rightarrow\) HCl còn dư, Fe phản ứng hết
\(\Rightarrow n_{H_2}=0,2mol\) \(\Rightarrow V_{H_2}=0,2\cdot22,4=4,48\left(l\right)\)
c) PTHH: \(HCl+NaOH\rightarrow NaCl+H_2O\)
Ta có: \(n_{HCl\left(dư\right)}=\dfrac{54}{365}\left(mol\right)=n_{NaOH}\)
\(\Rightarrow V_{NaOH}=\dfrac{\dfrac{54}{365}}{0,5}\approx0,3\left(l\right)=300\left(ml\right)\)
Theo giả thiết ta có : nCO2 = 6,72/22,4 = 0,3 (mol)
a) PTHH :
CO2+Ba(OH)2−>BaCO3↓+H2OCO2+Ba(OH)2−>BaCO3↓+H2O
0,3mol......0,3mol................0,3mol.........0,3mol
b) nồng độ mol của dd Ba(OH)2 đã dùng là :
CMBa(OH)2=0,30,6=0,5(M)CMBa(OH)2=0,30,6=0,5(M)
c) khối lượng kết tủa tạo thành là :
mBaCO3=0,3.197=59,1(g) Bn áp dụng làm nhé
a)
\(n_{CO_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)
PTHH: Na2CO3 + 2CH3COOH --> 2CH3COONa + CO2 + H2O
0,15<---------0,3<------------------------0,15
=> \(C\%_{dd.CH_3COOH}=\dfrac{0,3.60}{200}.100\%=9\%\)
b)
\(m_{dd.Na_2CO_3}=\dfrac{0,15.106.100}{15}=106\left(g\right)\)
c)
PTHH: 2CH3COOH + Ba(OH)2 --> (CH3COO)2Ba + 2H2O
0,3--------->0,15
=> \(V_{dd.Ba\left(OH\right)_2}=\dfrac{0,15}{0,5}=0,3\left(l\right)=300\left(ml\right)\)