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\(a)\\ n_{O_2} = \dfrac{6,72}{22,4} = 0,3(mol)\\ 4P + 5O_2 \xrightarrow{t^o} 2P_2O_5\\ \dfrac{n_P}{4} = 0,05 < \dfrac{n_{O_2}}{5} = 0,06\)
Do đó, Oxi dư.
\(n_{O_2\ pư} = \dfrac{5}{4}n_P = 0,25(mol)\\ \Rightarrow m_{O_2\ dư} = (0,3 - 0,25).32 = 1,6(gam)\\ b)\\ n_{P_2O_5} = \dfrac{n_P}{2} = 0,1(mol)\\ \Rightarrow m_{P_2O_5} = 0,1.142 = 14,2(gam)\)
\(n_P=\dfrac{6,2}{31}=0,2\left(mol\right)\\ n_{O_2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\\ PTHH:4P+5O_2\underrightarrow{t^o}2P_2O_5\\ LTL:\dfrac{0,2}{4}< \dfrac{0,4}{5}\Rightarrow O_2dư\)
\(n_{O_2\left(pư\right)}=\dfrac{5}{4}n_P=\dfrac{5}{4}.0,2=0,25\left(mol\right)\\ n_{O_2\left(dư\right)}=0,4-0,25=0,15\left(mol\right)\)
\(n_{P_2O_5\left(lt\right)}=\dfrac{1}{2}n_P=\dfrac{1}{2}.0,2=0,1\left(mol\right)\\ m_{P_2O_5\left(lt\right)}=0,1.142=14,2\left(g\right)\\ m_{P_2O_5\left(tt\right)}=0,1.142.80\%=11,36\left(g\right)\)
a) \(n_{Al}=\dfrac{12,15}{27}=0,45\left(mol\right)\)
\(n_{O_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)
PTHH: 4Al + 3O2 --to--> 2Al2O3
Xét tỉ lệ: \(\dfrac{0,45}{4}>\dfrac{0,3}{3}\)=> Al dư, O2 hết
PTHH: 4Al + 3O2 --to--> 2Al2O3
0,4<--0,3-------->0,2
=> \(m_{Al\left(dư\right)}=\left(0,45-0,4\right).27=1,35\left(g\right)\)
b) \(m_{Al_2O_3}=0,2.102=20,4\left(g\right)\)
Bài 1:
\(a,2Cu+O_2\underrightarrow{t^o}2CuO\)
b, \(n_{O_2}=\dfrac{1,12}{32}=0,035mol\)
\(n_{Cu}=\dfrac{6,4}{64}=0,1mol\)
\(\dfrac{0,1}{2}>\dfrac{0,035}{1}\) => Cu dư, O2 đủ
\(n_{Cu}\left(dư\right)=0,1-0,07=0,039\left(mol\right)\)
c, \(m_{CuO}=0,07.80=5,6g\)
Bài 2:
\(n_{Al}=\dfrac{13,5}{27}=0,5mol\)
\(n_{O_2}=\dfrac{6,67}{32}=0,21\left(mol\right)\)
\(4Al+3O_2\underrightarrow{t^o}2Al_2O_3\)
\(\dfrac{0,5}{4}>\dfrac{0,21}{3}\) => Al dư, O2 đủ
\(n_{Al_2O_3}=\dfrac{2}{3}.0,21=0,14\left(mol\right)\)
\(m_{Al_2O_3}=0,14.102=14,28g\)
a) $n_P = \dfrac{12,4}{31} = 0,4(mol) ; n_{O_2} = \dfrac{13,44}{22,4} = 0,6(mol)$
$4P + 5O_2 \xrightarrow{t^o} 2P_2O_5$
Ta thấy :
$n_P : 4 < n_{O_2} : 5$ nên $O_2$ dư
Điphotpho pentaoxit được tạo thành
$n_{P_2O_5} = \dfrac{1}{2}n_P = 0,2(mol)$
$m_{P_2O_5} = 0,2.142 = 28,4(gam)$
a)
\(n_{Al} = \dfrac{12,15}{27} = 0,45(mol)\\ n_{O_2} = \dfrac{6,72}{22,4} = 0,3(mol)\\ 4Al + 3O_2 \xrightarrow{t^o} 2Al_2O_3\\ \dfrac{n_{Al}}{4} = 0,1125 < \dfrac{n_{O_2}}{3} = 0,1\)
Do đó, Al dư.
\(n_{Al\ pư} = \dfrac{4}{3}n_{O_2} = 0,4(mol)\\ m_{Al\ dư} = (0,45-0,4).27 =1,35(gam)\)
b) Nhôm oxit được tạo thành.
\(n_{Al_2O_3} = \dfrac{2}{3}n_{O_2} = 0,2(mol)\\ \Rightarrow m_{Al_2O_3} = 0,2.102 = 20,4(gam)\)
\(n_{Al}=\dfrac{m}{M}=0,45\left(mol\right)\)
\(n_{O_2}=\dfrac{V}{22,4}=0,3\left(mol\right)\)
a, \(PTHH:4Al+3O_2\rightarrow2Al_2O_3\)
=> Sau phản ứng O2 hết, Al dư ( dư 0,05 mol )
=> \(m_{Aldu}=n.M=1,35\left(g\right)\)
b, Chất được tạo thành là Al2O3 .
Theo PTHH : \(n_{Al_2O_3}=0,2\left(mol\right)\)
\(\Rightarrow m_{Al_2O_3}=n.M=20,4\left(g\right)\)
Vậy ...
\(n_{Fe}=\dfrac{12.6}{56}=0.225\left(mol\right)\)
\(n_{O_2}=\dfrac{4.2}{22.4}=0.1875\left(mol\right)\)
\(3Fe+2O_2\underrightarrow{^{^{t^0}}}Fe_3O_4\)
\(3.........2\)
\(0.225......0.1875\)
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)
\(n_P = \dfrac{6,2}{31} = 0,2(mol) ; n_{O_2} = \dfrac{6,72}{22,4} = 0,3(mol)\\ 4P + 5O_2 \xrightarrow{t^o} 2P_2O_5\)
Ta thấy :
\(\dfrac{n_P}{4} = 0,05 < \dfrac{n_{O_2}}{5} = 0,06\) nên O2 dư.
\(n_{O_2\ pư} = \dfrac{5}{4}n_P = 0,25(mol)\\ \Rightarrow m_{O_2\ dư} = (0,3 - 0,25).32 = 1,6(gam)\)
b)
\(n_{P_2O_5} = \dfrac{n_P}{2} = 0,1(mol)$\\ m_{P_2O_5} = 0,1.142 = 14,2(gam)\)
\(n_P=\dfrac{6.2}{31}=0.2\left(mol\right)\)
\(n_{O_2}=\dfrac{6.72}{22.4}=0.3\left(mol\right)\)
\(4P+5O_2\underrightarrow{t^0}2P_2O_5\)
\(0.2.......0.25.........0.1\)
\(V_{O_2\left(dư\right)}=\left(0.3-0.25\right)\cdot22.4=1.12\left(l\right)\)
\(m_{P_2O_5}=0.1\cdot142=14.2\left(g\right)\)
PT: \(4P+5O_2\underrightarrow{t^o}2P_2O_5\)
a, Ta có: \(n_P=\dfrac{6,2}{31}=0,2\left(mol\right)\)
\(n_{O_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)
Xét tỉ lệ: \(\dfrac{0,2}{4}< \dfrac{0,3}{5}\), ta được O2 dư.
Theo PT: \(n_{O_2\left(pư\right)}=\dfrac{5}{4}n_P=0,25\left(mol\right)\)
\(\Rightarrow n_{O_2\left(dư\right)}=0,05\left(mol\right)\)
\(\Rightarrow V_{O_2\left(dư\right)}=0,05.22,4=1,12\left(l\right)\)
b, Theo PT: \(n_{P_2O_5}=\dfrac{1}{2}n_P=0,1\left(mol\right)\)
\(\Rightarrow m_{P_2O_5}=0,1.142=14,2\left(g\right)\)
Bạn tham khảo nhé!