Đáp án:

 Bài 5:

a/$⇒m_{Al} =5,4g\ và\ m_{FeO}=0,072g$

b/0,1M

Bài 6:

a/$\%V_{H_2}=66,67\%;\%V_{SO_2}=23,33\%$

b/ $m=23,8g$

Giải thích các bước giải:

 Bài 5: $n_{H_2}=\dfrac{0,672}{22,4}=0,03\ mol$

$FeO+H_SO_4\to FeSO_4+H_2O\\x\hspace{2cm}x\\2Al+3H_2SO_4\to Al_2(SO_4)_3+3H_2\\0,02\hspace{0,5cm}0,03\hspace{3cm}0,03$

a/$⇒m_{Al} =0,02×27=5,4g\ và\ 72x + 0,02×27=0,612\\⇒x=0,001\ mol;m_{FeO}=0,072g$

b/Theo PTHH

$n_{H_2SO_4}=x+0,03=0,031\ mol⇒C_{M}=\dfrac{0,031}{0,31}=0,1M$

Bài 6: $n_{H_2SO_4}=0,3\ mol$

$Fe+H_2SO_4\to FeSO_4+H_2\\x\hspace{2cm}x\hspace{3cm}x\\Na_2SO_3+H_2SO_4l\to Na_2SO_3+SO_2+H_2O\\y\hspace{3cm}y\hspace{4cm}y$

Theo bài ra, ta có hệ:

$\left \{ {{x+y=0,3} \atop {2x+64y=6,8}} \right.$ ⇔$\left \{ {{x=0,2} \atop {y=0,1}} \right.$ 

a/$n_{H_2}=x=0,2;\ n_{SO_2}=y=0,1⇒\\\%V_{H_2}=\dfrac{0,2}{0,2+0,1}×100\%=66,67\%\\⇒\%V_{SO_2}=23,33\ \%$

b/ $m=56x+126y=23,8g$