Đáp án + Gải thích các bước giải:
Ta có PTHH (1): $Mg+2HCl \xrightarrow{{}} MgCl_{2} + H_{2}$
a) $n_{Mg}=\dfrac{6}{24}=0,25(mol)$
Từ PTHH (1): $n_{H_{2}}=n_{Mg}=0,25(mol)$
$⇒V_{H_{2}(đktx)}=0,25 .22,4=5,6(l)$
b) Từ PTHH (1): $n_{HCl}=2n_{Mg}=2.0,25=0,5(mol)$
$⇒m_{HCl}=0,5.36,5=18,25(g)$
c) PTHH (2): $3H_{2}+ Fe_{2}O_{3}\xrightarrow{{}} 2Fe+3H_{2}O$
Tỉ lệ: \(\dfrac{n_{Fe_{2}O_{3}}{1}>\dfrac {n_{H_{2}}{3}\)
$⇒ Fe_{2}O_{3}$ dư, tính theo $H_{2}$
$n_{Fe}=\dfrac{2}{3}n_{H_{2}}=\dfrac{2}{3}.0,25=\dfrac{1}{6}(mol)$
$⇒m_{Fe}=\dfrac{1}{6}.56=9,33(g)$
$n_{Fe_{2}O_{3}}$ dư $=0,1- n_{Fe_{2}O_{3}}$pứ $= 0,1-\dfrac{1}{3}n_{H_{2}}=0,1- \dfrac{1}{12}=\dfrac{0,2}{12}(mol)$
$⇒m_{Fe_{2}O_{3}}$ dư $=\dfrac{0,2}{12}.160=2,67(g)$