Bài 4 :
\(n_P=\dfrac{9.3}{31}=0.3\left(mol\right)\)
\(4P+5O_2\underrightarrow{^{t^0}}2P_2O_5\)
\(0.3.....0.375.....0.15\)
\(m_{P_2O_5}=0.15\cdot142=21.3\left(g\right)\)
\(V_{O_2}=0.375\cdot22.4=8.4\left(l\right)\)
Bài 4:
Ta có: \(n_P=\dfrac{9,3}{31}=0,3\left(mol\right)\)
PT: \(4P+5O_2\underrightarrow{t^o}2P_2O_5\)
___0,3_0,375__0,15 (mol)
a, mP2O5 = 0,15.142 = 21,3 (g)
b, VO2 = 0,375.22,4 = 8,4 (l)
Bạn tham khảo nhé!
\(n_P=\dfrac{9,3}{31}=0,3\left(mol\right)\)
\(4P+2O_5\rightarrow2P_2O_5\)
0,3 0,15 0,15 (mol)
\(m_{P_2O_5}=0,15.142=21,3\left(g\right)\)
\(V_{O_2}=0,15.22,4=3,36\left(l\right)\)










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