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a, Đặt $\text{$n_{Mg}$ = a mol}$

$\rightarrow$ $\text{$n_{Al}$ = 2a mol}$

$\rightarrow$ $\text{24a + 27. 2a = 7,8 gam}$

$\rightarrow$ $\text{24a + 54a = 7,8 }$

$\rightarrow$ $\text{78a = 7,8}$

$\rightarrow$ $\text{a = 0,1 mol}$

$\rightarrow$ $\text{$n_{Mg}$ = 0,1 mol}$

$\rightarrow$ $\text{$n_{Al}$ = 0,1. 2 = 0,2 mol}$

Ta được:

* $\text{Số nguyên tử Al = 0,2. 6. $10^{23}$ = 1,2. $10^{23}$ (nguyên tử)}$

* $\text{Số nguyên tử Mg = 0,1. 6. $10^{23}$ = 0,6. $10^{23}$ (nguyên tử)}$

b, * $\text{$m_{Al}$ = n. M = 0,2. 27 = 5,4 gam}$

    * $\text{$m_{Mg}$ = n. M = 0,1. 24 = 2,4 gam}$

c, PTHH: $\text{2Al + 6HCl $\longrightarrow$ $2AlCl_3$ + $3H_2$}$

               $\text{Mg + 2HCl $\longrightarrow$ $MgCl_2$ + $H_2$}$

Theo PTHH $\rightarrow$ $\text{$n_{H_2}$ = $\dfrac{3}{2}$. $n_{Al}$ + $n_{mG}$ = 0,3 + 0,1 = 0,4 mol}$

$\rightarrow$ $\text{$V_{H_{2(đktx)}}$ = n. 22,4 = 0,4. 22,4 = 8,96 lít}$

$\text{#Kudo}$