Câu 1 :

$n_{NH_3}=54/17≈3,176mol$

$n_{O_2}=78,4/22,4=3,5mol$

a.PTHH :$

                      $4NH_3  +  3O_2\overset{t^o}\to 2N_2+6H_2O$

Theo pt :       4  mol       3 mol

Theo đbài : 3,176mol   3,5 mol

Ta có tỷ lệ :

$\dfrac{3,176}{4}<\dfrac{3,5}{3}$

⇒Sau pư O2 dư , NH3 phản ứng hết

Theo pt :

$n_{O_2\ pư}=3/4.n_{NH_3}=3/4.3,176=2,382mol$

$⇒n_{O_2\ dư}=3,5-2,382=1,118mol$

b.Theo pt :

$n_{N_2}=1/2.n_{NH_3}=1/2.3,176=1,588mol$

$⇒m_{N_2}=1,588.14=22,232g$

$n_{H_2O}=3/2.n_{NH_3}=3/2.3,176=4,764mol$

$⇒m_{H_2O}=4,764.18=85,752g$

Bài 2 :

$n_{O_2}=0,00896/22,4=0,0004mol$

$a.2KMnO_4\overset{t^o}\to K_2MnO_4+MnO_2+O_2$

Theo pt :

$n_{KMnO_4}=2.n_{O_2}=2.0,0004=0,0008mol$

$⇒m_{KMnO_4}=0,0008.158=0,1264g$

$b.2KClO_3\overset{t^o}\to 2KCl+3O_2$

Theo pt :

$n_{KClO_3}=2/3.n_{O_2}=2/3.0,0004=\dfrac{1}{375}mol$

$⇒m_{KClO_3}=\dfrac{1}{375}.122,5=\dfrac{49}{150}g$