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\(n_{FeCl_3}=\dfrac{48,75}{162,5}=0,3(mol)\\ 3NaOH+FeCl_3\to Fe(OH)_3\downarrow+3NaCl\\ \Rightarrow n_{Fe(OH)_3}=0,3(mol);n_{NaOH}=n_{NaCl}=0,9(mol)\\ a,m_{Fe(OH)_3}=0,3.107=32,1(g)\\ b,m_{dd_{NaOH}}=\dfrac{0,9.40}{10\%}=360(g)\\ c,C\%_{NaCl}=\dfrac{0,9.58,5}{360+48,75-32,1}.100\%=13,98\%\\ \)
\(d,2Fe(OH)_3+3H_2SO_4\to Fe_2(SO_4)_3+6H_2O\\ \Rightarrow n_{H_2SO_4}=0,45(mol)\\ \Rightarrow m_{dd_{H_2SO_4}}=\dfrac{0,45.98}{20\%}=220,5(g)\\ \Rightarrow V_{dd_{H_2SO_4}}=\dfrac{220,5}{1,14}=193,42(ml)\)
a) $n_{H_2SO_4} = \dfrac{490.10\%}{98} = 0,5(mol)$
$2NaOH + H_2SO_4 \to Na_2SO_4 + 2H_2O$
Theo PTHH :
$n_{NaOH} = 2n_{H_2SO_4} = 1(mol)$
$\Rightarrow m_{dd\ NaOH} = \dfrac{1.40}{20\%} = 200(gam)$
\(n_{H_2SO_4}=\dfrac{490.10\%}{98}=0,5\left(mol\right)\\ 2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O\\ n_{NaOH}=0,5.2=1\left(mol\right)\\ m_{ddNaOH}=\dfrac{1.40.100}{20}=200\left(g\right)\)
a) $n_{FeCl_3} = 0,5.3 = 1,5(mol) ; n_{NaOH} = 0,3.2 = 0,6(mol)$
$FeCl_3 + 3NaOH \to Fe(OH)_3 + 3NaCl$
Ta thấy :
$n_{FeCl_3} : 1 > n_{NaOH} : 3$ nên $FeCl_3 $ dư
$n_{Fe(OH)_3} = n_{NaOH} : 3 = 0,2(mol)$
$m_{Fe(OH)_3} = 0,2.107 = 21,4(gam)$
b) $2Fe(OH)_3 \xrightarrow{t^o} Fe_2O_3 + 3H_2O$
$n_{Fe_2O_3} = \dfrac{1}{2}n_{Fe(OH)_3} = 0,1(mol)$
$a = 0,1.160 = 16(gam)$
\(n_{CuSO_4}=\dfrac{200.16\%}{160}=0,2\left(mol\right)\)
PTHH :
\(CuSO_4+2NaOH\rightarrow Cu\left(OH\right)_2\downarrow+Na_2SO_4\)
0,2 0,4 0,2 0,2
\(m_{NaOH}=0,4.40=16\left(g\right)\)
\(m_{ddNaOH}=\dfrac{16.100}{10}=160\left(g\right)\)
\(c,m_{Na_2SO_4}=0,2.142=28,4\left(g\right)\)
\(m_{ddNa_2SO_4}=200+160-\left(0,2.98\right)=340,4\left(g\right)\)
\(C\%_{Na_2SO_4}=\dfrac{28,4}{240,4}.100\%\approx8,34\%\)
\(d,PTHH:\)
\(Cu\left(OH\right)_2\underrightarrow{t^o}CuO+H_2O\)
0,2 0,2
\(m_{CuO}=0,2.80=16\left(g\right)\)
a, \(CuSO_4+2NaOH\rightarrow Cu\left(OH\right)_2+Na_2SO_4\)
b, \(m_{CuSO_4}=200.16\%=32\left(g\right)\Rightarrow n_{CuSO_4}=\dfrac{32}{160}=0,2\left(mol\right)\)
\(n_{NaOH}=2n_{CuSO_4}=0,4\left(mol\right)\Rightarrow m_{ddNaOH}=\dfrac{0,4.40}{10\%}=160\left(g\right)\)
c, \(n_{Cu\left(OH\right)_2}=n_{Na_2SO_4}=n_{CuSO_4}=0,2\left(mol\right)\)
\(\Rightarrow C\%_{Na_2SO_4}=\dfrac{0,2.142}{200+160-0,2.98}.100\%\approx8,34\%\)
d, \(Cu\left(OH\right)_2\underrightarrow{t^o}CuO+H_2O\)
\(n_{CuO}=n_{Cu\left(OH\right)_2}=0,2\left(mol\right)\Rightarrow m_{CuO}=0,2.80=16\left(g\right)\)
\(m_{CH_3COOH}=12\%.100=12\left(g\right)\\ n_{CH_3COOH}=\dfrac{12}{60}=0,2\left(mol\right)\)
PTHH: CH3COOH + NaOH ---> CH3COONa + H2O
0,2--------->0,2------------>0,2
\(m_{NaOH}=0,2.40=8\left(g\right)\\ m_{ddNaOH}=\dfrac{8}{8,4\%}=\dfrac{2000}{21}\left(g\right)\\ m_{ddCH_3COONa}=\dfrac{2000}{21}+100=\dfrac{4100}{21}\left(g\right)\\ m_{CH_3COONa}=0,2.82=16,4\left(g\right)\\ C\%_{CH_3COONa}=\dfrac{16,4}{\dfrac{4100}{21}}.100\%=8,4\%\)
`=>` Gợi ý:
`CH3COOH + NaHCO3 => CH3COONa + CO2 + H2O`
`mCH3COOH = 100x12/100 = 12` (g)
`==> nCH3COOH = m/M = 12/60 = 0.2` (mol)
Theo pt: `=> nNaHCO3 = 0.2` (mol)
`==> mNaHCO3 = n.M = 0.2x84 =16.8` (g)
`==> mdd NaHCO3 = 16.8x100/8.4 = 200` (g)
Ta có: `nCH3COONa = 0.2` (mol)
PTHH: \(2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O\)
Ta có: \(n_{Na_2SO_4}=n_{H_2SO_4}=\dfrac{1}{2}n_{NaOH}=\dfrac{1}{2}\cdot\dfrac{10}{40}=0,125\left(mol\right)\) \(\Rightarrow\left\{{}\begin{matrix}m_{ddH_2SO_4}=\dfrac{0,125\cdot98}{10\%}=122,5\left(g\right)\\m_{Na_2SO_4}=0,125\cdot142=17,75\left(g\right)\end{matrix}\right.\)
\(\Rightarrow C\%_{Na_2SO_4}=\dfrac{17,75}{10+122,5}\cdot100\%\approx13,4\%\)
PTHH : FeCl3 + 3NaOH → Fe(OH)3 + 3NaCl
nFeCl3=\(\dfrac{26}{162,5}\)=0,16 mol . Theo tỉ lệ pt => nNaOH = 0,16.3 =0,48 mol.
<=> mNaOH = 0,48.40= 19,2 gam
C% =\(\dfrac{m_{\left(ct\right)}}{m_{\left(dd\right)}}.100\) => m dung dịch NaOH 10% = \(\dfrac{19,2.100}{10}\)= 192 gam
PT: \(FeCl_3+3NaOH\rightarrow3NaCl+Fe\left(OH\right)_{3\downarrow}\)
Ta có: \(n_{FeCl_3}=\dfrac{26}{162,5}=0,16\left(mol\right)\)
Theo PT: \(n_{NaOH}=3n_{FeCl_3}=0,48\left(mol\right)\)
\(\Rightarrow m_{NaOH}=0,48.40=19,2\left(g\right)\)
\(\Rightarrow m_{ddNaOH}=\dfrac{19,2.100}{10}=192\left(g\right)\)
Bạn tham khảo nhé!