a)
\(n_P = \dfrac{62}{31} = 2(mol)\\ 4P + 5O_2 \xrightarrow{t^o} 2P_2O_5\\ n_{O_2} = \dfrac{5}{4}n_P = 2,5(mol)\\ V_{O_2} = 2,5.22,4 = 56(lít)\\ V_{không\ khí} = \dfrac{56}{20\%} = 280(lít)\)
b)
\(n_P = \dfrac{31}{31} = 1(mol) ; n_{O_2} = \dfrac{23}{32} = 0,71875(mol)\\ \dfrac{n_P}{4} = 0,25 > \dfrac{n_{O_2}}{5} = 0,14375 \to P\ dư\\ n_{P\ pư} = \dfrac{4}{5}n_{O_2} = 0,575(mol)\\ m_{P\ dư} = 31 - 0,575.31 = 13,175(gam)\\ n_{P_2O_5} = \dfrac{2}{5}n_{O_2} = 0,2875(mol) \Rightarrow m_{P_2O_5} = 0,2875.142=40,825(gam)\)
Câu 1:
PTHH: \(4P+5O_2\underrightarrow{t^o}2P_2O_5\)
a) Ta có: \(n_P=\dfrac{62}{31}=2\left(mol\right)\) \(\Rightarrow n_{O_2}=2,5\left(mol\right)\)
\(\Rightarrow V_{O_2}=2,5\cdot22,4=56\left(l\right)\) \(\Rightarrow V_{kk}=\dfrac{56}{20\%}=280\left(l\right)\)
b) Ta có: \(\left\{{}\begin{matrix}n_P=\dfrac{31}{31}=1\left(mol\right)\\n_{O_2}=\dfrac{2}{32}=0,0625\left(mol\right)\end{matrix}\right.\)
Xét tỉ lệ: \(\dfrac{1}{4}>\dfrac{0,0625}{5}\) \(\Rightarrow\) Photpho còn dư, Oxi p/ư hết
\(\Rightarrow n_{P\left(dư\right)}=0,95\left(mol\right)\) \(\Rightarrow m_{P\left(dư\right)}=0,95\cdot31=29,45\left(g\right)\)
Theo PTHH: \(n_{P_2O_5}=0,025\left(mol\right)\) \(\Rightarrow m_{P_2O_5}=0,025\cdot142=3,55\left(g\right)\)
