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a) PTHH: \(FeCl_3+3NaOH\rightarrow3NaCl+Fe\left(OH\right)_3\downarrow\)
\(2Fe\left(OH\right)_3+3H_2SO_4\rightarrow Fe_2\left(SO_4\right)_3+6H_2O\)
b) Ta có: \(n_{FeCl_3}=0,3\cdot0,5=0,15\left(mol\right)\)
\(\Rightarrow n_{NaOH}=0,45mol\) \(\Rightarrow V_{ddNaOH}=\dfrac{0,45}{0,25}=1,8\left(l\right)\)
c) Theo PTHH: \(n_{NaCl}=n_{NaOH}=0,45mol\)
\(\Rightarrow C_{M_{NaCl}}=\dfrac{0,45}{2,1}\approx0,21\left(M\right)\)
(Coi như thể tích dd thay đổi không đáng kể)
d) Theo PTHH: \(n_{H_2SO_4}=\dfrac{3}{2}n_{Fe\left(OH\right)_3}=\dfrac{3}{2}n_{FeCl_3}=0,225mol\)
\(\Rightarrow m_{ddH_2SO_4}=\dfrac{0,225\cdot98}{20\%}=110,25\left(g\right)\)
\(\Rightarrow V_{ddH_2SO_4}=\dfrac{110,25}{1,14}\approx96,71\left(ml\right)\)
\(n_{H_2SO_4}=1.0,2=0,2(mol)\\ H_2SO_4+BaCl_2\to BaSO_4\downarrow+2HCl\\ \Rightarrow n_{BaSO_4}=n_{BaCl_2}=0,2(mol)\\ a,m_{BaSO_4}=0,2.233=46,6(g)\\ b,V_{dd_{BaCl_2}}=\dfrac{0,2}{1,5}\approx 0,13(l)\\ c,n_{HCl}=0,4(mol)\\ \Rightarrow C_{M_{HCl}}=\dfrac{0,4}{0,2+0,13}\approx 1,21M\)
\(d,\) Dd sau p/ứ là HCl nên làm quỳ tím hóa đỏ
\(n_{H_2SO_4}=0,2.1=0,2\left(mol\right)\\ H_2SO_4+BaCl_2\rightarrow BaSO_4+2HCl\\ n_{BaCl_2}=n_{BaSO_4}=n_{H_2SO_4}=0,2\left(mol\right)\\ n_{HCl}=2.0,2=0,4\left(mol\right)\\ a,m_{\downarrow}=m_{BaSO_4}=0,2.233=46,6\left(g\right)\\ b,V_{\text{dd}BaCl_2}=\dfrac{0,2}{1,5}=\dfrac{2}{15}\left(l\right)\\ c,C_{M\text{dd}HCl}=\dfrac{0,4}{\dfrac{2}{15}+0,2}=1,2\left(M\right)\\ d,V\text{ì}.c\text{ó}.\text{dd}.HCl\Rightarrow Qu\text{ỳ}.ho\text{á}.\text{đ}\text{ỏ}\)
\(n_{CuSO_4}=\dfrac{15,2}{160}=0,095mol\\ CuSO_4+2NaOH\rightarrow Cu\left(OH\right)_2+Na_2SO_4\)
0,095 0,19 0,095 0,095
\(m_{rắn}=m_{Cu\left(OH\right)_2}=0,095.98=9,31g\\ V_{ddNaOH}=\dfrac{0,19}{2}=0,095l\\ b)C_{M_{Na_2SO_4}}=\dfrac{0,095}{0,04+0,095}\approx0,7M\\ c)Cu\left(OH\right)_2\xrightarrow[t^0]{}CuO+H_2O\)
0,095 0,095
\(m_{rắn}=m_{CuO}=0,095.80=7,6g\)
\(a,PTHH:3NaOH+FeCl_3\rightarrow3NaCl+Fe\left(OH\right)_3\downarrow\\ 2Fe\left(OH\right)_3\rightarrow^{t^o}Fe_2O_3+3H_2O\uparrow\\ b,n_{FeCl_3}=1,5\cdot0,2=0,3\left(mol\right)\\ \Rightarrow n_{NaOH}=3n_{FeCl_3}=0,9\left(mol\right)\\ \Rightarrow V_{dd_{NaOH}}=\dfrac{0,9}{2}=0,45\left(l\right)\)
Theo đề: \(\left\{{}\begin{matrix}X:Fe\left(OH\right)_3\\A:NaCl\\Y:Fe_2O_3\end{matrix}\right.\)
Theo PT: \(n_{NaCl}=3n_{FeCl_3}=0,9\left(mol\right)\)
\(\Rightarrow C_{M_{NaCl}}=\dfrac{0,9}{0,45+0,2}\approx1,4M\)
\(c,\) Theo PT: \(n_{Fe\left(OH\right)_3}=n_{FeCl_3}=0,3\left(mol\right);n_{Fe_2O_3}=\dfrac{1}{2}n_{Fe\left(OH\right)_3}=0,15\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}m_X=m_{Fe\left(OH\right)_3}=0,3\cdot107=32,1\left(g\right)\\m_Y=m_{Fe_2O_3}=0,15\cdot160=24\left(g\right)\end{matrix}\right.\)
a, \(Na_2O+H_2O\rightarrow2NaOH\)
Ta có: \(n_{Na_2O}=\dfrac{15,5}{62}=0,25\left(mol\right)\)
Theo PT: \(n_{NaOH}=2n_{Na_2O}=0,5\left(mol\right)\)
\(\Rightarrow CM_{NaOH}=\dfrac{0,5}{0,5}=1\left(M\right)\)
b, \(2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O\)
Theo PT: \(n_{H_2SO_4}=\dfrac{1}{2}n_{NaOH}=0,25\left(mol\right)\)
\(\Rightarrow m_{ddH_2SO_4}=\dfrac{0,25.98}{20\%}=122,5\left(g\right)\)
\(\Rightarrow V_{ddH_2SO_4}=\dfrac{122,5}{1,14}\approx107,46\left(ml\right)\)
\(n_{FeCl_3}=\dfrac{48,75}{162,5}=0,3(mol)\\ 3NaOH+FeCl_3\to Fe(OH)_3\downarrow+3NaCl\\ \Rightarrow n_{Fe(OH)_3}=0,3(mol);n_{NaOH}=n_{NaCl}=0,9(mol)\\ a,m_{Fe(OH)_3}=0,3.107=32,1(g)\\ b,m_{dd_{NaOH}}=\dfrac{0,9.40}{10\%}=360(g)\\ c,C\%_{NaCl}=\dfrac{0,9.58,5}{360+48,75-32,1}.100\%=13,98\%\\ \)
\(d,2Fe(OH)_3+3H_2SO_4\to Fe_2(SO_4)_3+6H_2O\\ \Rightarrow n_{H_2SO_4}=0,45(mol)\\ \Rightarrow m_{dd_{H_2SO_4}}=\dfrac{0,45.98}{20\%}=220,5(g)\\ \Rightarrow V_{dd_{H_2SO_4}}=\dfrac{220,5}{1,14}=193,42(ml)\)