1)
$n_{Zn}$ = $\frac{19.5}{65}$ = 0.3 (mol)
a) $Zn_{}$ + $2HCl_{}$ → $ZnCl_{2}$ + $H_{2}↑$
b) Theo PTHH, ta có: $n_{H_{2}}$ = $n_{Zn}$ = 0.3 (mol)
⇒ $V_{2(đktc)}$ = 0.3 × 22.4 = 6.72 (l)
2) $n_{Fe_{2}O_{3}}$ = $\frac{19.2}{160}$ = 0.12 (mol)
$Fe_{2}$$O_{3}$ + $3H_{2}$ → $2Fe_{}$ + $3H_{2}$$O_{}$
Bđ : 0.12 0.3 (mol)
H=100% : 0.1 <--- 0.3 ----> 0.2 (mol)
H=80% : 0.08 0.24 0.16 (mol)
Sau pư : 0.04 0.06 0.16 (mol)
$m_{Fe}$ = 0.16 × 56 = 8.96 (g)
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