PTHH: \(4P+5O_2\underrightarrow{t^o}2P_2O_5\)
Ta có: \(n_P=\dfrac{6,2}{31}=0,2\left(mol\right)\)
a)
Theo PTHH: \(n_{O_2}=\dfrac{5}{4}n_P=0,25\left(mol\right)\)
\(\Rightarrow V_{O_2}=0,25.22,4=5,6\left(l\right)\)
\(\Rightarrow V_{kk}=5,6:20\%=28\left(l\right)\)
b)
Theo PTHH: \(n_{P_2O_5}=\dfrac{2}{5}n_{O_2}=0,1\left(mol\right)\)
\(\Rightarrow m_{P_2O_5}=0,1.142=14,2\left(g\right)\)
Vậy...
a) \(n_P=\dfrac{6,2}{31}=0,2\left(mol\right)\)
PTHH: \(4P+5O_2\underrightarrow{t^0}2P_2O_5\)
Theo PTHH: \(n_P:n_{O_2}=4:5\)
\(\Rightarrow n_{O_2}=n_P.\dfrac{5}{4}=0,2.\dfrac{5}{4}=0,25\left(mol\right)\)
\(\Rightarrow V_{O_2\left(đktc\right)}=0,25.22,4=5,6\left(l\right)\)
\(\Rightarrow V_{k^2}=\dfrac{5,6}{20\%}.100\%=28\left(l\right)\)
b) Theo PTHH: \(n_P:n_{P_2O_5}=4:2\)
\(\Rightarrow n_{P_2O_5}=n_P.\dfrac{1}{2}=0,2.\dfrac{1}{2}=0,1\left(mol\right)\)
\(\Rightarrow m_{P_2O_5}=0,1.142=14,2\left(g\right)\)
