Cho 4gam hỗn hợp gồm Mg và MgO tác dụng vừa đủ với dung dịch HCl 14,6 % sau phản ứng thoát ra 2,24 lít khí H2 ở điều kiện tiêu chuẩn . Hãy tính a. Thành phần phần trăm về khối lượng mỗi chất trong hỗn hợp b. Khối lượng HCl đã dùng

2 câu trả lời

Giải thích các bước giải:

nH2 = = 0,1 mol

Mg + 2HCl → MgCl2 + H2       (1)

MgO + 2HCl → MgCl2 + H2O   (2)

 = 24 x 0,1 = 2,4 g

=>  = 4 - 2,4 = 1,6 g

a.  %mMg =  x 100 = 60%

%mMgO = 100 - 60 = 40%

b. = 0,04 mol

Theo PTHH (2), có : 2nMgO=nHCl => nHCl = 0,08 mol

 = (0,2 + 0,08) x 36,5 = 10,22 g

 =  x 100 = 70g

 

Đáp án:

 

Giải thích các bước giải:

 $n_{H2}$ = $\frac{2,24}{22,4}$  = 0,1 mol

Mg + 2HCl → MgCl2 + H2

0,1  ←  0,2  ←   0,1  ←  0,1   mol

MgO + 2HCl → MgCl2 + H2O

$m_{Mg}$ = 24 x 0,1 = 2,4 g

⇒ $m_{MgO}$ = 4 - 2,4 = 1,6 g

a.  %Mg = $\frac{2,4}{4}$ x 100 = 60%

%MgO = 100 - 60 = 40%

b. $n_{MgO}$ = $\frac{1,6}{40}$ = 0,04 mol

MgO + 2HCl → MgCl2 + H2O

0,04 → 0,08    →   0,04 →  0,04  mol

$m_{HCl}$ = (0,2 + 0,08) x 36,5 = 10,22 g

$m_{ddHCl}$ = $\frac{10,22}{14,6}$ x 100 = 70g

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