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10.
\(n_{Fe}=\dfrac{16.8}{56}=0.3\left(mol\right)\)
\(4Fe+3O_2\underrightarrow{^{^{t^0}}}2Fe_2O_3\)
\(0.3.....0.225....0.15\)
\(V_{O_2}=0.225\cdot22.4=5.04\left(l\right)\)
\(Fe_2O_3+3H_2SO_4\rightarrow Fe_2\left(SO_4\right)_3+3H_2O\)
\(0.15...........0.45\)
\(m_{H_2SO_4}=0.45\cdot98=44.1\left(g\right)\)
11.
\(n_{Fe_2O_3}=\dfrac{48}{160}=0.3\left(mol\right)\)
\(Fe_2O_3+6HCl\rightarrow2FeCl_3+3H_2O\)
\(0.3...........1.8...........0.6\)
\(m_{FeCl_3}=0.6\cdot162.5=97.5\left(g\right)\)
\(m_{HCl}=1.8\cdot36.5=65.7\left(g\right)\)
Bài 10:
\(a,n_{Fe}=\dfrac{16,8}{56}=0,3(mol)\\ PTHH:4Fe+3O_2\xrightarrow{t^o}2Fe_2O_3\\ Fe_2O_3+3H_2SO_4\to Fe_2(SO_4)_3+3H_2O\\ \Rightarrow n_{O_2}=\dfrac{3}{4}n_{Fe}=0,225(mol)\\ \Rightarrow V_{O_2}=0,225.22,4=5,04(l)\\ b,n_{H_2SO_4}=3n_{Fe_2O_3}=3.\dfrac{1}{2}n_{Fe}=0,45(mol)\\ \Rightarrow m_{H_2SO_4}=0,45.98=44,1(g)\)
Bài 11:
\(a,n_{Fe_2O_3}=\dfrac{48}{160}=0,3(mol)\\ PTHH:Fe_2O_3+6HCl\to 2FeCl_3+3H_2O\\ \Rightarrow n_{FeCl_3}=2n_{Fe_2O_3}=0,6(mol)\\ \Rightarrow m_{FeCl_3}=0,6.162,5=97,5(g)\\ b,n_{HCl}=6n_{Fe_2O_3}=1,8(mol)\\ \Rightarrow m_{HCl}=1,8.36,5=65,7(g)\)
a) PTHH: \(4Fe+3O_2\underrightarrow{t^o}2Fe_2O_3\)
Ta có: \(n_{Fe}=\dfrac{m}{M}=\dfrac{16,8}{56}=0,3\left(mol\right)\)
PTPƯ: 4Fe + 3O2 \(\underrightarrow{t^o}\) 2Fe2O3
4 3 2
0,3 0,225 0,15
\(\Rightarrow V_{O_2}=n.24,79=0,225.24,79=5,57775\left(l\right)\)
\(4Al+3O_2\rightarrow\left(t^o\right)Al_2O_3\\ Al_2O_3+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2O\\ a,n_{Al}=\dfrac{8,1}{27}=0,3\left(mol\right)\\ n_{O_2}=\dfrac{3}{4}.0,3=0,225\left(mol\right)\\ V_{O_2\left(\text{đ}ktc\right)}=0,225.22,4=5,04\left(l\right)\\ b,n_{Al_2O_3}=\dfrac{1}{4}.0,3=0,075\left(mol\right)\\ n_{H_2SO_4}=3.0,075=0,225\left(mol\right)\\ m_{H_2SO_4}=m=0,225.98=22,05\left(g\right)\)
a. \(n_{Fe}=\dfrac{16,8}{56}=0,3\left(mol\right)\)
PTHH : 3Fe + 2O2 -to> Fe3O4
0,3 0,2 0,1
b. \(m_{Fe_3O_4}=0,1.232=23,2\left(g\right)\)
c. \(V_{O_2}=0,2.22,4=4,48\left(l\right)\)
a \(3Fe+2O_2\underrightarrow{t^o}Fe_3O_4\)
b \(\Rightarrow n_{Fe}=\dfrac{16,8}{56}=0,3mol\) \(\Rightarrow n_{Fe_3O_4}=\dfrac{1}{3}n_{Fe}=0,1mol\Rightarrow m_{Fe_3O_4}=0,1\cdot232=2,32g\)
c \(\Rightarrow n_{O_2}=\dfrac{2}{3}n_{Fe}=0,2mol\Rightarrow V_{O_2}=0,2\cdot22,4=4,48l\)
PTHH: \(3Fe+2O_2\underrightarrow{t^o}Fe_3O_4\)
Ta có: \(n_{Fe}=\dfrac{16,8}{56}=0,3\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}n_{O_2}=0,2mol\\n_{Fe_3O_4}=0,1mol\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{Fe_3O_4}=0,1\cdot232=23,2\left(g\right)\\V_{O_2}=0,2\cdot22,4=4,48\left(l\right)\end{matrix}\right.\)
\(n_{Fe}=\dfrac{8,4}{56}=0,15\left(mol\right)\\ pthh:3Fe+2O_2\underrightarrow{t^o}Fe_3O_{\text{4}}\)
0,15 0,1 0,05
\(m_{Fe_2O_4}=0,05.232=11,6\left(g\right)\\
V_{O_2}=0,1.11,4=2,24\left(l\right)\\
pthh:2KMnO_4\underrightarrow{t^o}K_2MnO_4+MnO_2+O_2\uparrow\)
0,2 0,1
\(m_{KMnO_4}=0,2.158=31,6\left(g\right)\)
\(n_{Fe}=\dfrac{8,4}{56}=0,15\left(mol\right)\\ pthh:3Fe+2O_2\underrightarrow{t^o}Fe_3O_4\)
0,15 0,1 0,05
\(m_{Fe_3O_{\text{ 4}}}=0,05.232=11,6\left(g\right)\\ V_{O_2}=0,1.22,4=2,24\left(l\right)\\ pthh:2KMnO_4\underrightarrow{t^o}K_2MnO_4+MnO_2+O_2\)
0,1 0,05
\(m_{KMnO_4}=0,1.158=15,8\left(g\right)\)
Bài 1:
PT: \(3Fe+2O_2\underrightarrow{t^o}Fe_3O_4\)
a, Ta có: \(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\)
\(n_{O_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)
Xét tỉ lệ: \(\dfrac{0,1}{3}< \dfrac{0,1}{2}\), ta được O2 dư.
Theo PT: \(n_{O_2\left(pư\right)}=\dfrac{2}{3}n_{Fe}=\dfrac{1}{15}\left(mol\right)\)
\(\Rightarrow n_{O_2\left(dư\right)}=0,1-\dfrac{1}{15}=\dfrac{1}{30}\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}m_{O_2\left(dư\right)}=\dfrac{1}{30}.32\approx1,067\left(g\right)\\V_{O_2\left(dư\right)}=\dfrac{1}{30}.2,24\approx0,746\left(l\right)\end{matrix}\right.\)
b, Theo PT: \(n_{Fe_3O_4}=\dfrac{1}{3}n_{Fe}=\dfrac{1}{30}\left(mol\right)\)
\(\Rightarrow m_{Fe_3O_4}=\dfrac{1}{30}.232\approx7,733\left(g\right)\)
Bài 2:
PT: \(3Fe+2O_2\underrightarrow{t^o}Fe_3O_4\)
a, Ta có: \(n_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right)\)
Theo PT: \(n_{Fe_3O_4}=\dfrac{1}{3}n_{Fe}=\dfrac{1}{15}\left(mol\right)\)
\(\Rightarrow m_{Fe_3O_4}=\dfrac{1}{15}.232\approx15,467\left(g\right)\)
b, Theo PT: \(n_{O_2}=\dfrac{2}{3}n_{Fe}=\dfrac{2}{15}\left(mol\right)\)
\(\Rightarrow V_{O_2}=\dfrac{2}{15}.22,4\approx2,9867\left(l\right)\)
c, PT: \(2N_2+5O_2\underrightarrow{t^o}2N_2O_5\)
Ta có: \(n_{N_2}=\dfrac{2,8}{28}=0,1\left(mol\right)\)
Xét tỉ lệ: \(\dfrac{0,1}{2}>\dfrac{\dfrac{2}{15}}{5}\), ta được N2 dư.
Theo PT: \(n_{N_2O_5}=\dfrac{2}{5}n_{O_2}=\dfrac{4}{75}\left(mol\right)\)
\(\Rightarrow m_{N_2O_5}=\dfrac{4}{75}.108=5,76\left(g\right)\)
Bạn tham khảo nhé!
Bài 1 :
\(n_{Fe}=\dfrac{5.6}{56}=0.1\left(mol\right)\)
\(n_{O_2}=\dfrac{2.24}{224}=0.1\left(mol\right)\)
\(3Fe+2O_2\underrightarrow{t^0}Fe_3O_4\)
\(Bđ:0.1......0.1\)
\(Pư:0.1.......\dfrac{1}{15}...\dfrac{1}{30}\)
\(Kt:0........\dfrac{1}{30}....\dfrac{1}{30}\)
\(V_{O_2\left(dư\right)}=\dfrac{1}{30}\cdot22.4=0.747\left(l\right)\)
\(m_{Fe_3O_4}=\dfrac{1}{30}\cdot232=7.73\left(g\right)\)
Bài 2 :
\(n_{Fe}=\dfrac{11.2}{56}=0.2\left(mol\right)\)
\(3Fe+2O_2\underrightarrow{t^0}Fe_3O_4\)
\(0.2.......0.3.......\dfrac{1}{15}\)
\(V_{O_2}=0.3\cdot22.4=6.72\left(l\right)\)
\(m_{Fe_3O_4}=\dfrac{1}{15}\cdot232=15.47\left(g\right)\)
\(n_{N_2}=\dfrac{2.8}{28}=0.1\left(mol\right)\)
\(2N_2+5O_2\underrightarrow{t^0}2N_2O_5\)
\(0.12......0.3........0.12\)
\(m_{N_2O_5}=0.12\cdot108=12.96\left(g\right)\)
gọi số mol Fe là x
pthh : 3Fe + 2O2 --t---> Fe3O4
x---->\(\dfrac{2}{3}\)x----------> \(\dfrac{x}{3}\)
=> mFe3O4= \(\dfrac{x}{3}\) . 232 = \(\dfrac{232x}{3}\) (G)
=> VO2 = \(\dfrac{2x}{3}\) . 22,4 = \(\dfrac{224x}{15}\) (L)
$a\big)$
$n_{Fe}=\frac{16,8}{56}=0,3(mol)$
$3Fe+2O_2\xrightarrow{t^o}Fe_3O_4$
Theo PT: $n_{Fe_3O_4}=\frac{1}{3}n_{Fe}=0,1(mol)$
$\to m_{Fe_3O_4}=0,1.232=23,2(g)$
$b\big)$
Theo PT: $n_{O_2}=\frac{2}{3}n_{Fe}=0,2(mol)$
$\to V_{O_2}=0,2.22,4=4,48(l)$
$\to V_{kk}=4,48.5=22,4(l)$
$c\big)$
$2KMnO_4\xrightarrow{t^o}K_2MnO_4+MnO_2+O_2$
Theo PT: $n_{KMnO_4}=2n_{O_2}=0,4(mol)$
$\to m_{KMnO_4(dùng)}=\frac{0,4.158}{80\%}=79(g)$
a, \(n_{Fe}=\dfrac{16,8}{56}=0,3\left(mol\right)\)
PTHH: 3Fe + 2O2 ---to→ Fe3O4
Mol: 0,3 0,2 0,1
\(m_{Fe_3O_4}=0,1.232=23,2\left(g\right)\)
b, \(V_{O_2}=0,2.22,4=4,48\left(l\right)\Rightarrow V_{kk}=4,48.5=22,4\left(l\right)\)
c,
PTHH: 2KMnO4 ---to→ K2MnO4 + MnO2 + O2
Mol: 0,4 0,2
\(m_{KMnO_4\left(lt\right)}=0,4.158=63,2\left(g\right)\)
\(\Rightarrow m_{KMnO_4\left(tt\right)}=\dfrac{63,2}{80\%}=79\left(g\right)\)
a, \(n_{Fe}=\frac{16,8}{56}=0,3\left(mol\right)\)
\(3Fe+2O_2=Fe_3O_4\)
\(\Rightarrow n_{O_2}=0,2\left(mol\right);n_{Fe_3O_4}=0,1\left(mol\right)\)
\(V_{O_2}=0,2.22,4=4,481\)
b, \(Fe_3O_4+4H_2SO_4=FeSO_4+Fe_2\left(So_4\right)_3+4H_2O\)
\(\Rightarrow n_{H_2SO_4}=4n_{Fe_3O_4}=0,4\left(mol\right)\)
\(\Rightarrow m_{H_2SO_4}=0,4.98=39,2\left(g\right)\)