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nFe = 11.2/56 = 0.2 (mol)
Fe + H2SO4 => FeSO4 + H2
0.2____0.2_______0.2___0.2
mH2SO4 = 0.2*98 = 19.6 (g)
mdd H2SO4 = 19.6*100/10 = 196 (g)
m dd sau phản ứng = 11.2 + 196 - 0.4 = 206.8 (g)
mFeSO4 = 0.2*152 = 30.4 (g)
C% FeSO4 = 30.4/206.8 * 100% = 14.7%
Vdd H2SO4 = mdd H2SO4 / D = 196 / 1.14 = 171.9 (ml)
CM FeSO4 = 0.2 / 0.1719 = 1.16 M
a)
$Fe + H_2SO_4 \to FeSO_4 + H_2$
b) $n_{Fe} = \dfrac{14}{56} = 0,25(mol)$
Theo PTHH : $n_{H_2SO_4} = n_{Fe} = 0,25(mol)$
$\Rightarrow m_{dd\ H_2SO_4} = \dfrac{0,25.98}{10\%} = 245(gam)$
c)
$n_{H_2} = n_{Fe} = 0,25(mol)$
Sau phản ứng, $m_{dd} = 14 + 245 - 0,25.2 = 258,5(gam)$
$C\%_{FeSO_4} = \dfrac{0,25.152}{258,5}.100\% = 14,7\%$
\(n_{FeCl_3}=\dfrac{48,75}{162,5}=0,3(mol)\\ 3NaOH+FeCl_3\to Fe(OH)_3\downarrow+3NaCl\\ \Rightarrow n_{Fe(OH)_3}=0,3(mol);n_{NaOH}=n_{NaCl}=0,9(mol)\\ a,m_{Fe(OH)_3}=0,3.107=32,1(g)\\ b,m_{dd_{NaOH}}=\dfrac{0,9.40}{10\%}=360(g)\\ c,C\%_{NaCl}=\dfrac{0,9.58,5}{360+48,75-32,1}.100\%=13,98\%\\ \)
\(d,2Fe(OH)_3+3H_2SO_4\to Fe_2(SO_4)_3+6H_2O\\ \Rightarrow n_{H_2SO_4}=0,45(mol)\\ \Rightarrow m_{dd_{H_2SO_4}}=\dfrac{0,45.98}{20\%}=220,5(g)\\ \Rightarrow V_{dd_{H_2SO_4}}=\dfrac{220,5}{1,14}=193,42(ml)\)
PTHH: \(Fe+2HCl\rightarrow FeCl_2+H_2\uparrow\)
Ta có: \(n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}n_{Fe}=n_{FeCl_2}=n_{H_2}=0,2\left(mol\right)\\n_{HCl}=0,4\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{Fe}=0,2\cdot56=11,2\left(g\right)\\m_{FeCl_2}=0,2\cdot127=25,4\left(g\right)\\m_{ddHCl}=\dfrac{0,4\cdot36,5}{10\%}=146\left(g\right)\\m_{H_2}=0,2\cdot2=0,4\left(g\right)\end{matrix}\right.\)
Mặt khác: \(m_{dd}=m_{Fe}+m_{ddHCl}-m_{H_2}=156,8\left(g\right)\) \(\Rightarrow C\%_{FeCl_2}=\dfrac{25,4}{156,8}\cdot100\%\approx16,2\%\)
Đáp án:
m =32,4g
mddH2SO4 = 49g
Giải thích các bước giải:
a) MgCO3 + H2SO4 → MgSO4 + H2O +CO2 ↑
MgSO4 + 2NaOH → Mg(OH)2 + Na2SO4
$Mg{(OH)_2}\buildrel {to} \over
\longrightarrow MgO + {H_2}O$
b) nCO2 = 2,24 : 22,4 = 0,1mol
nMgCO3 = nCO2 = 0,1 mol
nMgO = 12:40=0,3mol
nMgSO4 = nMgO - nMgCO3 = 0,3 - 0,1 = 0,2mol
m = mMgCO3 + mMgSO4
= 0,1 .84+0,2.120=32,4g
nH2SO4 = nCO2 = 0,1 mol
mH2SO4 = 0,1.98=9,8g
mddH2SO4 = 9,8:20.100=49g
chúc bạn học tốt
Ta có: \(n_{Fe_3O_4}=\dfrac{2,32}{232}=0,01\left(mol\right)\)
a. PTHH: Fe3O4 + 4H2SO4 ---> FeSO4 + Fe2(SO4)3 + 4H2O
Theo PT: \(n_{H_2SO_4}=4.n_{Fe_3O_4}=4.0,01=0,04\left(mol\right)\)
=> \(m_{H_2SO_4}=0,04.98=3,92\left(g\right)\)
Theo đề, ta có: \(C_{\%_{H_2SO_4}}=\dfrac{3,92}{m_{dd_{H_2SO_4}}}.100\%=20\%\)
=> \(m_{dd_{H_2SO_4}}=19,6\left(g\right)\)
b. Ta có: \(m_{dd_{SauPỨ}}=2,32+19,6=21,92\left(g\right)\)
Theo PT: \(n_{FeSO_4}=n_{Fe_2\left(SO_4\right)_3}=n_{Fe_3O_4}=0,01\left(mol\right)\)
=> \(m_{FeSO_4}=0,01.152=1,52\left(g\right)\)
\(m_{Fe_2\left(SO_4\right)_3}=0,01.400=4\left(g\right)\)
=> \(m_{SauPỨ}=1,52+4=5,52\left(g\right)\)
=> \(C_{\%_{SauPỨ}}=\dfrac{5,52}{21,92}.100\%=25,18\%\)
nFe = 11.2/56 = 0.2 (mol)
Fe + H2SO4 => FeSO4 + H2
0.2____0.2_______0.2___0.2
mH2SO4 = 0.2*98 = 19.6 (g)
mdd H2SO4 = 19.6*100/10 = 196 (g)
m dd sau phản ứng = 11.2 + 196 - 0.4 = 206.8 (g)
mFeSO4 = 0.2*152 = 30.4 (g)
C% FeSO4 = 30.4/206.8 * 100% = 14.7%
Vdd H2SO4 = mdd H2SO4 / D = 196 / 1.14 = 171.9 (ml)
CM FeSO4 = 0.2 / 0.1719 = 1.16 M