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Cứ 1 mol F e 2 O 3 có 2 mol Fe
Vậy 0,2 mol F e 2 O 3 có x? mol Fe
- Trong 0,125 mol phân tử PbO có 0,125 mol nguyên tử Pb.
Khối lượng của Pb = mPb=nPb.MPb=0,125.207=25,875(g)
Trong 0,35 mol phân tử CuO có 0,35 mol nguyên tử Cu.
Khối lượng của nguyên tử Cu: M C u = n C u . M C u =0,35.6=22,4(g)
nH2 = \(\dfrac{3,6}{22,4}\) = 0,16 mol
nFe2O3 = \(\dfrac{32}{160}\) = 0,2 mol
3H2 + Fe2O3 -> 2Fe + 3H2O
0,16(hết);0,2(dư) ->0,106 ->0,16
mH2O = 0,106 . 18 = 1,908 g
mFe = 0,16.56 = 8,96 g
a)3H2 + Fe2O3 -> 2Fe + 3H2O
b)nH2 = 3,6/22,4 \(\simeq\)0,16 (mol)
nFe2O3 = 32/160 = 0,2 (mol)
ta có: 0,16/3 < 0,2/1 => H2 hết, Fe2O3 dư
theo Pt ta có: nFe = 2/3.nH2=2/3.0,16\(\simeq\)0,1( mol)
n H2O=nH2=0,16 (mol)
=> m Fe=0,1.56=5,6(g)
mH2O=0,16.18=2,88(g)
\(n_{HCl}=\dfrac{14,6\%.450}{36,5}=1,8\left(mol\right)\\ n_{Fe_2O_3}=\dfrac{32}{160}=0,2\left(mol\right)\\ Fe_2O_3+6HCl\rightarrow2FeCl_3+3H_2O\\ Vì:\dfrac{1,8}{6}>\dfrac{0,2}{1}\\ \Rightarrow HCldư\\ a.n_{FeCl_3}=0,2.2=0,4\left(mol\right)\\ m_{FeCl_3}=162,5.0,4=65\left(g\right)\\ b.n_{HCl\left(dư\right)}=1,8-6.0,2=0,6\left(mol\right)\\ m_{HCl\left(dư\right)}=0,6.36,5=21,9\left(g\right)\\ c.m_{ddsau}=32+450=482\left(g\right)\\ C\%_{ddHCl\left(dư\right)}=\dfrac{21,9}{482}.100\approx4,544\%\\ C\%_{ddFeCl_3}=\dfrac{65}{482}.100\approx13,485\%\)
\(n_{Fe2O3}=\dfrac{32}{160}=0,2\left(mol\right)\)
\(m_{ct}=\dfrac{14,6.450}{100}=65,7\left(g\right)\)
\(n_{HCl}=\dfrac{65,7}{36,5}=1,8\left(mol\right)\)
Pt : \(Fe_2O_3+6HCl\rightarrow2FeCl_3+3H_2O|\)
1 6 2 3
0,2 1,8 0,4
a) Lập tỉ số so sánh : \(\dfrac{0,2}{1}< \dfrac{1,8}{6}\)
⇒ Fe2O3 phản ứng hết , HCl dư
⇒ Tính toán dựa vào số mol của Fe2O3
\(n_{FeCl3}=\dfrac{0,2.2}{1}=0,4\left(mol\right)\)
⇒ \(m_{FeCl3}=0,4.162,5=65\left(g\right)\)
b) \(n_{HCl\left(dư\right)}=1,8-\left(0,2.6\right)=0,6\left(mol\right)\)
⇒ \(m_{HCl\left(dư\right)}=0,6.36,5=21,9\left(g\right)\)
c) \(m_{ddspu}=32+450=482\left(g\right)\)
\(C_{FeCl3}=\dfrac{65.100}{482}=13,48\)0/0
\(C_{HCl\left(dư\right)}=\dfrac{21,9.100}{482}=4,54\)0/0
Chúc bạn học tốt
\(n_{Fe_2O_3}=\dfrac{32}{160}=0,2\left(mol\right)\)
\(m_{H_2SO_4}=196.40\%=78,4\left(g\right)\Rightarrow n_{H_2SO_4}=\dfrac{78,4}{98}=0,8\left(mol\right)\)
PTHH: Fe2O3 + 3H2SO4 → Fe2(SO4)3 + 3H2O
Mol: 0,2 0,6 0,2
Ta có: \(\dfrac{0,2}{1}< \dfrac{0,8}{3}\) ⇒ Fe2O3 hết, H2SO4 dư
mdd sau pứ = 32 + 196 = 228 (g)
\(C\%_{ddFe_2\left(SO_4\right)_3}=\dfrac{0,2.400.100\%}{228}=35,09\%\)
\(C\%_{ddH_2SO_4dư}=\dfrac{\left(0,8-0,6\right).98.100\%}{228}=8,596\%\)
a)\(n_{Fe_2O_3}=0,2\left(mol\right)\)
PT:\(Fe_2O_3+6HCl\rightarrow2FeCl_3+3H_2O\)
\(0,2\) \(1,2\) \(0,4\)
\(\Rightarrow n_{FeCl_3}=0,4\left(mol\right)\)
\(\Rightarrow m_{FeCl_3}=65\left(g\right)\)
b) \(n_{HCl}=\dfrac{218.30\%}{35,5+1}=\dfrac{654}{365}\left(mol\right)\)
Từ PT \(\Rightarrow\)\(n_{HClpư}=1,2\left(mol\right)\)
\(\Rightarrow n_{HCldư}=\dfrac{654}{365}-1,2=\dfrac{216}{365}\left(mol\right)\)
\(\Rightarrow m_{HCldư}=21,6\left(g\right)\)
\(m_{dd}=32+218=250\left(g\right)\)
\(C\%_{FeCl_3}=\dfrac{65}{250}.100\%=26\left(\%\right)\)
\(C\%_{HCldu}=\dfrac{21,6}{250}.100\%=8,64\%\)
Fe2O3+3CO--->2Fe+3CO2
n Fe2O3=32/160=0,2(mol)
n Fe=22,4/56=0,4(mol)
Theo pthh
n Fe=2n Fe2O3 =0,4(mol)
-->Fe2O3 phản ứng hết
Theo pthh
n CO=3n Fe2O3=0,6(mol)
V CO=0,6.22,4=13,44(l)
PTHH: \(Fe_2O_3+6HCl\rightarrow2FeCl_3+3H_2O\)
\(2FeCl_3+Cu\rightarrow CuCl_2+2FeCl_2\)
Ta có: \(\left\{{}\begin{matrix}n_{FeCl_3}=2n_{Fe_2O_3}=2\cdot\dfrac{16}{160}=0,2\left(mol\right)\\n_{Cu}=\dfrac{32}{64}=0,5\left(mol\right)\end{matrix}\right.\)
Xét tỉ lệ: \(\dfrac{0,2}{2}< \dfrac{0,5}{1}\) \(\Rightarrow\) Cu còn dư
\(\Rightarrow\left\{{}\begin{matrix}n_{CuCl_2}=0,1\left(mol\right)\\n_{FeCl_2}=0,2\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{CuCl_2}=0,1\cdot135=13,5\left(g\right)\\m_{FeCl_2}=0,2\cdot127=25,4\left(g\right)\end{matrix}\right.\)
nFe2O3= 32/160=0.2mol
nCuO=28/80=0.35mol
mPbO=223*0.125=27.875g
số ptu Fe2O3=6*10^23*0.2=1.2*10^23(ptu)
=> số ntu Fe = .2*10^23 * 2= 2.4*10^23(ntu);
mFe = 0.2* 56* 2 = 22.4g
%Fe=(22.4/32)*100%=70%
sô ptu PbO = 6*10^23 * 0.125 = 7.5*10^22(ptu)
=>số ntu Pb=7.5*10^22(ntu)
mPb=0.125*207=25.875g
%Pb=(25.875/27.875)*100%=92.83%
số ptu CuO=6*10^23*0.35=2.1*10^23(ptu)
=>số ntu Cu = 2.1*10^23(ntu)
mCu= 0.35*64=22.4g
%Cu=(22.4/28)*100%=80%
\\tham khảo//
1) 32g Fe2O3
\(n_{Fe_2O_3}=\dfrac{32}{160}=0,2\left(mol\right)\)
Ta có: \(n_{Fe}=2n_{Fe_2O_3}=2\times0,2=0,4\left(mol\right)\)
\(\Rightarrow m_{Fe}=0,4\times56=22,4\left(g\right)\)
\(\%m_{Fe}=\dfrac{22,4}{32}\times100\%=70\%\)
Ta có: \(n_O=3n_{Fe_2O_3}=3\times0,2=0,6\left(mol\right)\)
\(\Rightarrow m_O=0,6\times16=9,6\left(g\right)\)
\(\%m_O=\dfrac{9,6}{32}\times100\%=30\%\)
2) 28g CuO
\(n_{CuO}=\dfrac{28}{80}=0,35\left(mol\right)\)
Ta có:\(n_{Cu}=n_O=n_{CuO}=0,35\left(mol\right)\)
\(\Rightarrow m_{Cu}=0,35\times64=22,4\left(g\right)\)
\(\%m_{Cu}=\dfrac{22,4}{28}\times100\%=80\%\)
\(m_O=0,35\times16=5,6\left(g\right)\)
\(\%m_O=\dfrac{5,6}{28}\times100\%=20\%\)
3) 45g Fe(OH)2
\(n_{Fe\left(OH\right)_2}=\dfrac{45}{90}=0,5\left(mol\right)\)
Ta có: \(n_{Fe}=n_{Fe\left(OH\right)_2}=0,5\left(mol\right)\)
\(\Rightarrow m_{Fe}=0,5\times56=28\left(g\right)\)
\(\%m_{Fe}=\dfrac{28}{45}\times100\%=62,22\%\)
Ta có: \(n_O=n_H=2n_{Fe\left(OH\right)_2}=2\times0,5=1\left(mol\right)\)
\(\Rightarrow m_O=1\times16=16\left(g\right)\)
\(\%m_O=\dfrac{16}{45}\times100\%=35,56\%\)
\(m_H=1\times1=1\left(g\right)\)
\(\%m_H=\dfrac{1}{45}\times100\%=2,22\%\)
4) 0,125 mol PbO
\(m_{PbO}=0,125\times223=27,875\left(g\right)\)
Ta có: \(n_{Pb}=n_O=n_{PbO}=0,125\left(mol\right)\)
\(\Rightarrow m_{Pb}=0,125\times207=25,875\left(g\right)\)
\(\%m_{Pb}=\dfrac{25,875}{27,875}\times100\%=92,83\%\)
\(m_O=0,125\times16=2\left(g\right)\)
\(\%m_O=\dfrac{2}{27,875}\times100\%=7,17\%\)