$4Al+3O_2\xrightarrow{t^o}2Al_2O_3$

$n_{O_2}=\dfrac{5,04}{22,4}=0,225(mol)$

Theo PT: $n_{Al_2O_3}=\dfrac{2}{3}n_{O_2}=0,15(mol)$

$\Rightarrow m_{Al_2O_3}=0,15.102=15,3(g)$

$\Rightarrow m_{Al(trong X)}=17,2-15,3=1,9(g)$

Theo PT: $n_{Al}=\dfrac{4}{3}n_{O_2}=0,3(mol)$

$\Rightarrow m_{Al(p/ứ)}=0,3.27=8,1(g)$

$\Rightarrow \Sigma m_{Al}=8,1+1,9=10(g)$

$\to m=10$

$\%m_{Al(trong X)}=\dfrac{1,9}{17,2}.100\%\approx 11,05\%$

$\%m_{Al_2O_3}=100-11,05=88,95\%$

$4Al+3O_2\xrightarrow{t^o}2Al_2O_3$

$n_{O_2}=\dfrac{V}{22,4}=\dfrac{6,72}{22,4}=0,3(mol)$

Theo PT: $n_{Al}=\dfrac{4}{3}n_{O_2}=0,4(mol)$

$\Rightarrow m_{Al}=n.M=0,4.27=10,8(g)$

$a)n_{Fe}=\dfrac{42}{56}=0,75(mol)$

$Fe_2O_3+3H_2\xrightarrow{t^o}2Fe+3H_2O$

$\Rightarrow n_{Fe_2O_3}=0,5n_{Fe}=0,375(mol)$

$\Rightarrow m_{Fe_2O_3}=0,375.160=60(g)$

$b)n_{H_2O}=1,5n_{Fe}=1,125(mol)$

$\Rightarrow m_{H_2O}=1,125.18=20,25(g)$

$\Rightarrow n-1\in Ư(-16)=\left\{-16;-8;-4;-2;-1;1;2;4;8;16\right\}$

$\Rightarrow n\in \left\{-15;-7;-3;-1;0;2;3;5;9;17\right\}$

có mác nào màu khác không ạ cho e đăng kí với :>

đỉnh đỉnh :3

đệ tử :>