4P+5O2-to>2P2O5
0,2---0,25------0,1 mol
n P=\(\dfrac{6,2}{31}\)=0,2 mol
=>VO2=0,25.22,4=5,6l
=>Vkk=5,6.5=28l
c)
2KClO3-to>2KCl+3O2
\(\dfrac{1}{6}\)--------------------------0,25 mol
H=90%
m KClO3 thực tế=\(\dfrac{1}{6}\).122,5.\(\dfrac{100}{90}\)=22,685g
\(n_P=\dfrac{6,2}{31}=0,2\left(mol\right)\\ a,PTHH:4P+5O_2\rightarrow\left(t^o\right)2P_2O_5\\ b,n_{O_2}=\dfrac{5}{4}.n_P=\dfrac{5}{4}.0,2=0,25\left(mol\right)\\ \Rightarrow V_{O_2\left(đktc\right)}=0,25.22,4=5,6\left(l\right)\\ V_{kk}=5.V_{O_2}=5.5,6=28\left(l\right)\\ c,2KClO_3\rightarrow\left(t^o\right)2KCl+3O_2\uparrow\\ n_{KClO_3\left(LT\right)}=\dfrac{2}{3}.n_{O_2}=\dfrac{2}{3}.0,25=\dfrac{1}{6}\left(mol\right)\\ \Rightarrow n_{KClO_3\left(TT\right)}=\dfrac{1}{6}.110\%=\dfrac{11}{60}\left(mol\right)\\ \Rightarrow m_{KClO_3\left(tt\right)}=\dfrac{11}{60}.122,5=\dfrac{539}{24}\left(g\right)\)
Bài 2:
a. PTHH: \(4P+5O_2\rightarrow^{t^0}2P_2O_5\)
b. \(n_P=\dfrac{6,2}{31}=0,2\left(mol\right)\)
\(4P+5O_2\rightarrow^{t^0}2P_2O_5\)
4 5 2
0,2 0,25 0,1
\(V_{O_2}=0,1.24,79=2,479\left(l\right)\)
\(V_{KK}=V_{O_2}.5=2,479.5=12,395\left(l\right)\).
c. PTHH: \(2KClO_3\rightarrow^{t^0}2KCl+3O_2\).
2 2 3
0,2 0,2 0,1
-\(m_{KClO_3}=M.n=122,5.0,2=24,5\left(g\right)\)
-Vì lượng hao hụt là 10% nên khối lượng của KClO3 trên thực tế là:
\(24,5.\dfrac{100-10}{100}=22,05\left(g\right)\)
