$n_{Al}=\dfrac{5,4}{27}=0,2mol \\a.PTHH : \\2Al+3H_2SO_4\to Al_2(SO_4)_3+3H_2↑ \\b.n_{H_2}=\dfrac{6,72}{22,4}=0,3mol \\Theo\ pt : \\n_{Al_2(SO_4)_3}=\dfrac{1}{3}.n_{H_2}=\dfrac{1}{3}.0,3=0,1mol \\⇒m_{Al_2(SO_4)_3}=0,1.342=34,2g \\n_{H_2SO_4}=n_{H_2}=0,3mol \\⇒m_{H_2SO_4}=0,3.98=29,4g \\c.V_{O_2}=\dfrac{5,6}{5}=1,12l \\⇒n_{O_2}=\dfrac{1,12}{22,4}=0,05mol$

                     $2H_2  +  O_2\overset{t^o}\to 2H_2O$

Theo pt :       2 mol       1 mol

Theo đbài :  0,3 mol     0,05 mol

Tỉ lệ : $\dfrac{0,3}{2}>\dfrac{0,05}{1}$

⇒Sau phản ứng H2 dư

Theo pt :

$n_{H_2O}=2.n_{O_2}=2.0,05=0,1mol$

$⇒m_{H_2O}=0,1.18=1,8g$

Đáp án:

Giải thích các bước giải:

 a) `2Al + 3H_2SO_4 -> Al_2(SO_4)_3 + 3H_2\uparrow `

b) `n_{Al}=\frac{5,4}{27}=0,2\ mol`

`n_{Al_2(SO_4)_3}=\frac{1}{2}.n_{Al}=0,1\ mol`

`m_{Al_2(SO_4)_3}=0,1.342=34,2\ gam`

`n_{H_2SO_4}=\frac{3}{2}.n_{Al}=0,3\ mol`

`m_{H_2SO_4}=0,3.98=29,4\ gam`

c) `2H_2+O_2 \rightarrow^{t^{0}} 2H_2O`

`n_{O_2}=\frac{5,6:5}{22,4}=0,05\ mol`

`n_{H_2}=\frac{6,72}{22,4}=0,3\ mol`

`\frac{n_{H_2}}{2}>\frac{n_{O_2}}{1}`

`⇒ H_2` dư, `O_2` hết

`n_{H_2O}=2.n_{O_2}=0,1\ mol`

`m_{H_2O}=0,1.18=1,8\ gam`