Đáp án:

 b) mZnO=2,43 gam

c) V O2=0,336 lít

d) mKMnO4=4,74 gam

Giải thích các bước giải:

 nZn=1,95/65=0,03 mol

2Zn + O2 --> 2ZnO

Ta có: nZnO=nZn=0,03 mol -> mZnO=0,03.(65+16)=2,43 gam

nO2=1/2n Zn=0,03/2=0,015 mol -> V O2=0,015.22,4=0,336 lít

2KMnO4 ---> K2MnO4 + MnO2 + O2

Ta có: nKMnO4=2nO2=0,015.2=0,03 mol -> mKMnO4=0,03.(39+55+16.4)=4,74 gam

a,

$2Zn+O_2\buildrel{{t^o}}\over\to 2ZnO$

b,

$n_{Zn}=\dfrac{1,95}{65}=0,03 mol$

$\Rightarrow n_{ZnO}=0,03 mol$

$m_{ZnO}=0,03.81=2,43g$

c,

$n_{O_2}=\dfrac{1}{2}.0,03=0,015 mol$

$\Rightarrow V_{O_2}=0,015.22,4=0,336l$

d,

$2KMnO_4\buildrel{{t^o}}\over\to K_2MnO_4+MnO_2+O_2$

$\Rightarrow n_{KMnO_4}=0,015.2=0,03 mol$

$m_{KMnO_4}=0,03.158=4,74g$