\(a) 4P + 5O_2 \xrightarrow{t^o} 2P_2O_5\\ b) n_P = \dfrac{12,4}{31} = 0,4(mol)\\ n_{O_2} = \dfrac{5}{4}n_P = 0,5(mol)\\ V_{O_2} = 0,5.11,2 = 11,2(lít)\\\ c) 2KClO_3 \xrightarrow{t^o} 2KCl + 3O_2\\ n_{KClO_3} = \dfrac{2}{3}n_{O_2} = \dfrac{1}{3}(mol)\\ m_{KClO_3}= \dfrac{1}{3}.122,5 = 40,83(gam)\)
\(n_P=\dfrac{12.4}{31}=0.4\left(mol\right)\)
\(4P+5O_2\underrightarrow{t^0}2P_2O_5\)
\(0.4........0.5\)
\(V_{O_2}=0.5\cdot22.4=11.2\left(l\right)\)
\(2KClO_3\underrightarrow{t^0}2KCl+3O_2\)
\(\dfrac{1}{3}.................0.5\)
\(m_{KClO_3}=\dfrac{1}{3}\cdot122.5=40.83\left(g\right)\)
a) PTHH : 4P+5O2→ 2P2O5
b) nP= \(\dfrac{12,4}{31}\)=0,4(mol)
→ nO2= 0,5(mol)
vO2= 0,5 . 22,4=11,2(l)
c) PTHH : 2KClO3→ 2KCl + 3O2
nKClO3= \(\dfrac{1}{3}\)(mol)
→ mKClO3= \(\dfrac{1}{3}\). 122,5= 40,83(g)
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