PT: \(4P+5O_2\underrightarrow{t^o}2P_2O_5\)
Ta có: \(n_P=\dfrac{3,1}{31}=0,1\left(mol\right)\)
a, Theo PT: \(n_{O_2}=\dfrac{5}{4}n_P=0,125\left(mol\right)\)
\(\Rightarrow V_{O_2}=0,125.22,4=2,8\left(l\right)\)
b, Theo PT: \(n_{P_2O_5}=\dfrac{1}{2}n_P=0,05\left(mol\right)\)
\(\Rightarrow m_{P_2O_5}=0,05.142=7,1\left(g\right)\)
Bạn tham khảo nhé!
\(n_P=\dfrac{3.1}{31}=0.1\left(mol\right)\)
\(4P+5O_2\underrightarrow{t^0}2P_2O_5\)
\(0.1...0.125....0.05\)
\(V_{O_2}=0.125\cdot22.4=2.8\left(l\right)\)
\(m_{P_2O_5}=0.05\cdot142=7.1\left(g\right)\)
PTHH : 4P + 5\(O_2\) -> 2\(P_2O_5\)
a) nP = m / M = 0.1(mol)
có : nP = 5/4 n\(O_2\)
-> n\(O_2\) = 0.125 (mol)
-> V\(O_2\) (đktc) = n.22.4 = 2.8 (L)
b) có : nP = 1/2 n\(P_2O_5\)
-> n\(P_2O_5\) = 0.05 (mol)
-> m\(P_2O_5\) = n.M = 7.1 (g)
a.nP=3,1/31=0,1 mol
nO=5/4.0,1=0,125 mol
VO=0,125.22,4=2.8 lít(đktc)
b.nP2O5=0.1/2=0,05mol
mP2O5=0,05.142=7,1g
