Hòa tan hoàn toàn 4,8 gam Mg vào dung dịch HCl 10,95% a) viết phương trình hóa học b) tính khối lượng dung dịch HCl cần dùng c) tính thể tích khí sinh ra ở đktc d) tính nồng độ % các chất trong dung dịch tạo thành

2 câu trả lời

$\text{#Đỉnh cao tri thức ( ĐCTT )}$

Đáp án:

`a)`

PTHH : `Mg + 2HCl -> MgCl_2 + H_2`

`b)`

`n_{Mg} = (4,8)/(24) = 0,2` `mol`

`n_{HCl} = 2 . n_{Mg} = 0,4` `mol`

`m_{HCl} = 0,4 . 36,5 = 14,6` `gam`

`m_{dd_(HCl)} = (14,6)/(10,95%) . 100% = 133,33` `gam`

`c)`

`n_{H_2} = n_{Mg} = 0,2` `mol`

`V_{H_2} = 0,2 . 22,4 = 4,48` `l`

`d)`

`n_{MgCl_2} = n_{Mg} = 0,2` `mol`

`m_{MgCl_2} = 0,2 . 95 = 19` `gam`

`m_{dd_(pứ)} = 4,8 + 133,33 - 0,2 . 2 = 137,73` `gam`

`C%_{MgCl_2} = (19)/(137,73) . 100% = 13,79%`

 

a, PTHH : Mg + 2HCl => MgCl2 + H2

b, nMg = 4,8 /24 = 0,2 (mol)

Theo PTHH ta có :

nHCl = 2.0,2 = 0,4 (mol ) 

mct HCl = 0,4 . 36,5 = 14,6 (g)

mddHCl = 14,6/10,95 % = 133,3 (g)

c, Theo PTHH ta có nH2=nMg = 0,2 (mol)

VH2 (đktc) = 0,2 . 22,4=4,48 (l)

d, nMgCl2= nMg= 0,2 (mol)

=> mMgCl2 = 0,2 . 95= 19(g)

mH2 = 0,2 .2 = 0,4 (g)

mdd sau pứ :  (133,3 + 4,8 )- 0,4 =  137,7 (g)

C% MgCl2 = (19/137 ,7 ) .100%= 13,798 %

~Chúc e học tốt~

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