\(a.pthh:4P+5O_2\overset{t^o}{--->}2P_2O_5\)
b. Ta có: \(n_P=\dfrac{3,1}{31}=0,1\left(mol\right)\)
Theo pt: \(n_{O_2}=\dfrac{5}{4}.n_P=\dfrac{5}{4}.0,1=0,125\left(mol\right)\)
\(\Rightarrow V_{O_2}=0,125.22,4=2,8\left(lít\right)\)
c. \(V_{kk}=V_{O_2}.5=2,8.5=14\left(lít\right)\)
d. Theo pt: \(n_{P_2O_5}=\dfrac{1}{2}.n_P=\dfrac{1}{2}.0,1=0,05\left(mol\right)\)
\(\Rightarrow m_{P_2O_5}=0,05.142=7,1\left(g\right)\)
a,\(n_P=\dfrac{3,1}{31}=0,1\left(mol\right)\)
pthh 4P + 5O2 → (to) 2P2O5
0,1 0,125 0,05
b, \(V_{O_2}=0,125.22,4=2,8\left(l\right)\\ V_{kk}=2,8.5=14\left(l\right)\\ m_{P_2O_5}=0,05.142=7,1\left(g\right)\)
PTHH : \(4P+5O_2\left(t^o\right)->2P_2O_5\) (1)
\(n_P=\dfrac{m}{M}=\dfrac{3,1}{31}=0,1\left(mol\right)\)
Từ (1) -> \(\dfrac{5}{4}n_P=n_{O_2}=0,125\left(mol\right)\)
-> \(V_{O_2\left(dktc\right)}=n.22,4=0,125.22,4=2,8\left(l\right)\)
\(V_{kk}=5.V_{O_2}=5.2,8=14\left(l\right)\)
Từ (1) -> \(\dfrac{1}{2}n_P=n_{P_2O_5}=0,05\left(mol\right)\)
-> \(m_{P_2O_5}=n.M=0,05.\left(31.2+16.5\right)=7,1\left(g\right)\)
\(a,n_P=\dfrac{3,1}{31}=0,1\left(mol\right)\\ 4P+5O_2\rightarrow\left(t^o\right)2P_2O_5\\ b,n_{O_2}=\dfrac{5}{4}.n_P=\dfrac{5.0,1}{4}=0,125\left(mol\right)\\ \Rightarrow V_{O_2\left(đktc\right)}=0,125.22,4=2,8\left(l\right)\\ c,V_{kk}=5.V_{O_2}=5.2,8=14\left(l\right)\\ d,n_{P_2O_5}=\dfrac{2}{4}.n_P=\dfrac{4.0,1}{2}=0,05\left(mol\right)\\ \Rightarrow m_{P_2O_5}=142.0,05=7,1\left(g\right)\)
