Bài 1 :
$n_{Zn}=\dfrac{3,25}{65}=0,05mol \\m_{HCl}=50.7,3\%=3,65g \\⇒n_{HCl}=\dfrac{3,65}{36,5}=0,1mol \\PTHH :$
$Zn + 2HCl\to ZnCl_2+H_2↑$
Theo pt : 1 mol 2 mol
Theo đbài : 0,05 mol 0,1 mol
Tỉ lệ : $\dfrac{0,05}{1}=\dfrac{0,1}{2}$
⇒Phản ứng xảy ra hoàn toàn
$a.Theo\ pt : \\n_{H_2}=n_{Zn}=0,05mol \\⇒V_{H_2}=0,05.22,4=1,12l \\b.Theo\ pt: \\n_{ZnCl_2}=n_{Zn}=0,05mol \\⇒m_{ZnCl_2}=0,05.136=6,8(g) \\m_{dd\ spu}=3,25+50-0,05.2=53,15(g) \\⇒C\%_{ZnCl_2}=\dfrac{6,8}{53,15}.100\%=12,79\%$
Bài 2 :
a.-Kim loại :
Zn : kẽm
K : kali
Ca : Canxi
-Oxit bazo :
Al2O3 : nhôm oxit
K2O : kali oxit
BaO : bari oxit
HgO : thuỷ ngân II oxit
-Oxit axit :
SO3 : lưu huỳnh trioxit
P2O5 : điphotpho pentaoxit
-Bazo :
KOH : kali hidroxit
-Muối :
FeS : sắt II sunfua
BaSO4 : bari sunfat
Bài 3 :
$n_{Mg}=\dfrac{7,2}{24}=0,3mol \\PTHH : \\Mg+H_2SO_4\to MgSO_4+H_2↑ \\a.Theo\ pt : \\n_{H_2SO_4}=n_{Mg}=0,3mol \\⇒m_{H_2SO_4}=0,3.98=29,4g \\b.Theo\ pt : \\n_{MgSO_4}=n_{H_2}=n_{Mg}=0,3mol \\⇒m_{MgSO_4}=0,3.120=36(g) \\m_{dd\ H_2SO_4}=\dfrac{29,4}{19,6\%}=150(g) \\m_{dd\ spu}=7,2+150-0,3.2=156,6g \\⇒C\%_{MgSO_4}=\dfrac{36}{156,6}.100\%=22,99\%$