a) \(n_{C_4H_{10}}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\)
PTHH: 2C4H10 + 13O2 --to--> 8CO2 + 10H2O
0,4---->2,6---------->1,6------->2
=> m = 1,6.44 = 70,4 (g)
b) \(V_{O_2}=2,6.22,4=58,24\left(l\right)\)
c) \(n_P=\dfrac{9,1}{31}=\dfrac{91}{310}\left(mol\right)\)
PTHH: 4P + 5O2 --to--> 2P2O5
Xét tỉ lê \(\dfrac{\dfrac{91}{310}}{4}< \dfrac{2,6}{5}\) => P hết, O2 dư
\(m_{P_2O_5}=\dfrac{91}{620}.142=\dfrac{6461}{310}\left(g\right)\)
Tên sản phẩm: Điphotpho pentaoxit