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Ta có: \(n_P=\dfrac{12,4}{31}=0,4\left(mol\right)\)
PT: \(4P+5O_2\underrightarrow{t^o}2P_2O_5\)
_____0,4____0,5_____0,2 (mol)
a, \(m_{P_2O_5}=0,2.142=28,4\left(g\right)\)
b, \(V_{O_2}=0,5.22,4=11,2\left(l\right)\)
a, \(4P+5O_2\underrightarrow{t^o}2P_2O_5\)
b, \(n_P=\dfrac{12,4}{31}=0,4\left(mol\right)\)
Theo PT: \(n_{P_2O_5}=\dfrac{1}{2}n_P=0,2\left(mol\right)\Rightarrow m_{P_2O_5}=0,2.142=28,4\left(g\right)\)
c, \(n_{O_2}=\dfrac{5}{4}n_P=0,5\left(mol\right)\)
\(\Rightarrow V_{O_2}=0,5.22,4=11,2\left(l\right)\)
d, \(2KClO_3\underrightarrow{t^o}2KCl+3O_2\)
Theo PT: \(n_{KClO_3}=\dfrac{2}{3}n_{O_2}=\dfrac{1}{3}\left(mol\right)\Rightarrow m_{KClO_3}=\dfrac{1}{3}.122,5=\dfrac{245}{6}\left(g\right)\)
a. \(n_{KMnO_4}=\dfrac{47.4}{158}=0,3\left(mol\right)\)
PTHH : 2KMnO4 ---to----> K2MnO4 + MnO2 + O2
0,3 0,15
\(V_{O_2}=0,15.22,4=3,36\left(l\right)\)
b. PTHH : 4Al + 3O2 -> 2Al2O3
0,2 0,15
\(m_{Al}=0,2.27=5,4\left(g\right)\)
\(4P+5O_2\underrightarrow{t^o}2P_2O_5\)
\(nO_2=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)
\(\Rightarrow nP_2O_5=\dfrac{2}{5}nO_2=\dfrac{2}{5}.0,1=0,04\left(mol\right)\)
\(\Rightarrow mP_2O_5=0,04.\left(31.2+16.5\right)=5,68\left(g\right)\)
c/
pthh: \(2KMnO_4\rightarrow K_2MnO_4+MnO_2+O_2\uparrow\)
\(nO_2=0,1\left(mol\right)\)
\(\Rightarrow nKMnO_4=\dfrac{2}{1}.nO_2=\dfrac{2}{1}.0,1=0,2\left(mol\right)\)
\(\Rightarrow mKMnO_4=0,2.\left(39+55+16.4\right)=31,6\left(g\right)\)
nO2 = 2,479/24,79 = 0,1 (mol)
PTHH: 4P + 5O2 -> (t°) 2P2O5
Mol: 0,08 <--- 0,1 ---> 0,04
mP2O5 = 0,05 . 142 = 5,68 (g)
\(4P+5O_2\underrightarrow{t^o}2P_2O_5\)
\(nO_2=\dfrac{2,479}{24,79}=0,1\left(mol\right)\)
\(\Rightarrow nP_2O_5=\dfrac{2}{5}.nO_2=\dfrac{2}{5}.0,1=0,04\left(mol\right)\)
\(mP_2O_5=0,04.142=5,68\left(g\right)\)