PTHH: \(4P+5O_2\underrightarrow{t^o}2P_2O_5\)
Ta có: \(n_P=\dfrac{6,2}{31}=0,2\left(mol\right)\)
\(n_{O_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)
Theo PTHH ta có: \(\dfrac{0,2}{4}=0,05< \dfrac{0,3}{5}=0,06\)
\(a.\Rightarrow O_2dư\), \(P\) hết. Vậy ta tính theo \(n_P\)
Theo PTHH ta có:
\(n_{O_2\left(pư\right)}=\dfrac{0,2.5}{4}=0,25\left(mol\right)\)
\(n_{O_2\left(dư\right)}=0,3-0,25=0,05\left(mol\right)\)
\(\Rightarrow m_{O_2}=0,05.32=1,6\left(g\right)\)
b. Theo PTHH ta có: \(n_{P_2O_5}=\dfrac{0,2.2}{4}=0,1\left(mol\right)\)
\(\Rightarrow\) Khối lượng \(P_2O_5\) thu được là:
\(m_{P_2O_5}=0,1.142=14,2\left(g\right)\)
\(n_P=\dfrac{6,2}{31}=0,2\left(mol\right)\)
\(n_{O_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)
PTHH: 4P + 5O2 \(\underrightarrow{to}\) 2P2O5
Ban đầu: 0,2......0,3....................(mol)
phản ứng: 0,2.......0,25..................(mol)
sau phản ứng: 0.......0,05..→....0,1...(mol)
a) Sau phản ứng O2 dư
\(m_{O_2}dư=0,05\times32=1,6\left(g\right)\)
b) \(m_{P_2O_5}=0,1\times142=14,2\left(g\right)\)