Câu 9:
a) $Na_{2}$$CO_{3}$ + $2HCl_{}$ → $2NaCl_{}$ + $H_{2}$$O_{}$ + $CO_{2}↑$
b) $Ba(OH)_{2}$ + $H_{2}$$SO_{4}$ → $BaSO_{4}↓$ + $2H_{2}$$O_{}$
c) $ZnO_{}$ + $2HCl_{}$ → $ZnCl_{2}$ + $H_{2}$$O_{}$
d) $K_{2}$$CO_{3}$ + $2HCl_{}$ → $2KCl_{}$ + $H_{2}$$O_{}$ + $CO_{2}↑$
e) $Ba(OH)_{2}$ + $H_{2}$$SO_{4}$ → $BaSO_{4}↓$ + $2H_{2}$$O_{}$
f) $FeO_{}$ + $2HCl_{}$ → $FeCl_{2}$ + $H_{2}$$O_{}$
Câu 10:
$n_{BaCl_{2}}$ = $\frac{20.8}{208}$ = 0.1 (mol)
a) $BaCl_{2}$ + $H_{2}$$SO_{4}$ → $BaSO_{4}↓$ + $2HCl_{}$
b) Theo PTHH, ta có: $n_{BaSO_{4}}$ = $BaCl_{2}$ = 0.1 mol
⇒ $m_{BaSO_{4}}$ = 0.1 × 233 = 23.3 (g)
#HONKAI IMPACT
