Đáp án:

1/

4R+3O2=2R2O3

 2/15    0,1

nO2=0,1mol

MR=$\frac{13}{2/15}$ =97,5

=>Kl là........(xem lại giùm đề)

2/

4Al+3O2=2Al2O3

 x     0,75x

2Mg+O2=2MgO

y     0,5y

nO2=0,25mol

27x+24y=10,2

0,75x+0,5y=0,25

=>x=0,2 y=0,2

a/ %Al=$\frac{0,2.27}{10,2}$ .100=52,94%

%Mg=47,06%

b/

2KClO3=2KCl+3O2

$\frac{1}{6}$                   0,25

mKClO3=$\frac{1}{6}$ .122,5=20,42g

$1/$

$n_{O2}=\dfrac{2,24}{22,4}=0,1mol$

$PTHH:$

$4R+3O2→2R2O3$

Theo pt:

$n_{R}=4/3.n_{O2}=4/3.0,1=\dfrac{2}{15}mol$

$⇒M_R=\dfrac{13}{\dfrac{2}{15}}=97,5 (?)$

$2/$

$n_{O2}=5,6/22,4=0,25mol$

Gọi $n_{Al}=x(mol);n_{Mg}=y(mol) (x,y>0)$

$4Al+3O2\overset{t^o}{\rightarrow}2Al2O3(1)$

$2Mg+O2\overset{t^o}{\rightarrow}2MgO(2)$

Ta có :

$m_{Al}+m_{Mg}=27x+24y=10,2g$

$n_{O2}=n_{O2(1)}+n_{O2(2)}=0,75x+0,5y=0,25$

Ta có hệ pt:

$\left\{\begin{matrix} 27x+24y=10,2& \\ 0,75x+0,5y=0,25& \end{matrix}\right.$

giải hệ ta được:

$\left\{\begin{matrix} x=0,2 & \\ y=0,2& \end{matrix}\right.$

$a/$

$m_{Al}=0,2.27=5,4g$

$⇒\%m_{Al}=\dfrac{5,4.100\%}{10,2}=52,94\%$

$⇒\%m_{Mg}=100\%-52,94\%=47,06\%$

$b/$

$2KClO3\overset{t^o}{\rightarrow}2KCl+3O2$

Theo pt:

$n_{KClO3}=2/3.n_{O2}=2/3.0,25=\dfrac{1}{6}mol$

$⇒m_{KClO3}=\dfrac{1}{6}.122,5=\dfrac{245}{12}g$