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\(a,PTHH:3Fe+2O_2\xrightarrow{t^o}Fe_3O_4\\ b,n_{Fe}=\dfrac{5,6}{56}=0,1(mol);n_{O_2}=\dfrac{3,2}{32}=0,1(mol)\)
Vì \(\dfrac{n_{Fe}}{3}<\dfrac{n_{O_2}}{2}\) nên \(O_2\) dư
\(n_{O_2(dư)}=0,1-0,1.\dfrac{2}{3}=0,033(mol)\\ \Rightarrow m_{O_2(dư)}=0,033.32=1,056(mol)\\ c,n_{Fe_3O_4}=\dfrac{1}{3}n_{Fe}=0,033(mol)\\ \Rightarrow m_{Fe_3O_4}=0,033.232=7,656(g)\)
a) \(3Fe+2O_2-t^o->Fe_3O_4\)
b) \(n_{Fe}=\frac{5,6}{56}=0,1\left(mol\right)\)
Theo pthh : \(n_{Fe_3O_4}=\frac{1}{3}n_{Fe_3O_4}=\frac{0,1}{3}\left(mol\right)\)
=> \(m_{Fe_3O_4}=232\cdot\frac{0,1}{3}\approx7,73\left(g\right)\)
c) Theo pthh : \(n_{O2\left(pứ\right)}=\frac{2}{3}n_{Fe}=\frac{0,2}{3}\left(mol\right)\)
=> \(n_{O2\left(can.dung\right)}=\frac{0,2}{3}\div100\cdot120=0,08\left(mol\right)\)
=> \(V_{O2\left(can.dung\right)}=0,08\cdot22,4=1,792\left(l\right)\)
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 :
\(n_{Fe}=\dfrac{11.2}{56}=0.2\left(mol\right)\)
\(3Fe+2O_2\underrightarrow{t^0}Fe_3O_4\)
\(0.2.......0.3.......\dfrac{1}{15}\)
\(V_{O_2}=0.3\cdot22.4=6.72\left(l\right)\)
\(m_{Fe_3O_4}=\dfrac{1}{15}\cdot232=15.47\left(g\right)\)
\(n_{N_2}=\dfrac{2.8}{28}=0.1\left(mol\right)\)
\(2N_2+5O_2\underrightarrow{t^0}2N_2O_5\)
\(0.12......0.3........0.12\)
\(m_{N_2O_5}=0.12\cdot108=12.96\left(g\right)\)
a)
\(3Fe + 2O_2 \xrightarrow{t^o} Fe_3O_4\)
b)
Ta có :
\(n_{Fe} = \dfrac{8,4}{56} = 0,15(mol)\\ n_{O_2} = \dfrac{96}{32} = 3(mol)\)
Ta thấy : \(\dfrac{n_{Fe}}{3} = 0,05 < \dfrac{n_{O_2}}{2} = 1,5\) do đó O2 dư.
Theo PTHH :
\(n_{O_2\ pư} = \dfrac{2}{3}n_{Fe} = 0,1(mol)\\ \Rightarrow n_{O_2\ dư} = 3 - 0,1 = 2,9(mol)\\ \Rightarrow m_{O_2\ dư} = 92,8(gam)\)
c)
\(n_{Fe_3O_4} = \dfrac{1}{3}n_{Fe} = 0,05(mol)\\ \Rightarrow m_{Fe_3O_4} = 0,05.232 = 11,6(gam)\)
\(a)PTHH:FeCl_3+2O_2\xrightarrow[]{t^o}Fe_3O_4\)
mol 1 2 1
mol
\(b)\)Số mol \(FeCl_3\) là: \(n_{FeCl_3}=\dfrac{m_{FeCl_3}}{M_{FeCl_3}}=\dfrac{8,4}{162,5}=0,052\left(mol\right)\)
Số mol \(O_2\) là: \(n_{O_2}=\dfrac{m_{O_2}}{M_{O_2}}=\dfrac{96}{32}=3\left(mol\right)\)
Lập tỉ lệ: \(\dfrac{1}{0,052}>\dfrac{2}{3}\Rightarrow FeCl_3dư\)
Số mol \(FeCl_3\) phản ứng là:
Từ PTHH\(\Rightarrow\) \(n_{FeCl_3}=\dfrac{0,052\times3}{3}=0,035\left(mol\right)\)
Số mol \(FeCl_3\) dư là: \(n_{FeCl_3dư}=n_{FeCl_3đầu}-n_{FeCl_3p/ứng}=0,052-0,035=0,018\left(mol\right)\)
Khối lượng \(FeCl_3\) dư là: \(m_{FeCl_3dư}=n_{FeCl_3dư}\times M_{FeCl_3}=0,018\times162,5=2,925\left(g\right)\)
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 \(\Rightarrow n_{O_2}=\dfrac{2}{3}n_{Fe}=0,2mol\Rightarrow V_{O_2}=0,2\cdot22,4=4,48l\)
3Fe + 2O2 → Fe3O4
Theo pt : 3 2 1 mol
Theo đề bài : 0,2 0,3 0,2/3
a.
Ta có tỉ lệ \(\dfrac{0,2}{3}< \dfrac{0,3}{2}\) nên Fe phản ứng hết , oxi dư số mol sắt từ thu được tính theo Fe
b. nFe3O4 = 0,2/3 mol ==> m Fe3O4 = 0,2 /3 .232 = 15,47 gam
\(n_{Fe}=\dfrac{12.6}{56}=0.225\left(mol\right)\)
\(n_{O_2}=\dfrac{4.2}{22.4}=0.1875\left(mol\right)\)
\(3Fe+2O_2\underrightarrow{^{^{t^0}}}Fe_3O_4\)
\(3.........2\)
\(0.225......0.1875\)
Lập tỉ lệ : \(\dfrac{0.225}{3}< \dfrac{0.1875}{2}\Rightarrow O_2dư\)
\(m_{O_2\left(dư\right)}=\left(0.1875-0.225\cdot\dfrac{2}{3}\right)\cdot32=1.2\left(g\right)\)
\(m_{Fe_3O_4}=\dfrac{0.225}{3}\cdot232=17.4\left(g\right)\)
a) Sắt+oxi---> sắt từ
b) Theo định luật bảo toàn khối lượng
m\(_{Fe3O4}=m_{Fe}+m_{O2}\)
=1,68+0,64=2,32(g)