Đáp án:
b/ 0,1g
c/ 202,7g
d/ 3,75%
Giải thích các bước giải:
a/ $Fe+H_2SO_4\to FeSO_4+H_2$
b/
$n_{Fe}=\dfrac{2,8}{56}=0,05\ mol$
$m_{H_2SO_4}=\dfrac{35}{100}.200=70g\\⇒n_{H_2SO_4}=\dfrac{70}{98}=\dfrac{5}{7}\ mol$
Ta có: $0,05< \dfrac{5}{7}$ ⇒ Fe hết, axit dư
$⇒n_{H_2}=n_{Fe}=0,05 ⇒m_{H_2}=0,05.2=0,1g$
c/ $m_{dd\ sau\ pư}=m_{Fe}+m_{dd\ H_2SO_4}-m_{H_2}\\⇔m_{dd\ sau\ pư}=2,8+200-0,1=202,7g$
d/ Theo PTHH
$n_{FeSO_4}=n_{Fe}=0,05\ mol ⇒m_{FeSO_4}=0,05.152=7,6g\\⇒C\%=\dfrac{7,6}{202,7}.100\%=3,75\%$