hoà tan 17,775g hon hop kim loai fe, al,zn vao dung dich chua hcl sau phan ung hu duoc 6,72 lit h2 o dktc tinh khoi luong muoi thu duoc
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hoà tan 17,775g hon hop kim loai fe, al,zn vao dung dich chua hcl sau phan ung hu duoc 6,72 lit h2 o dktc tinh khoi luong muoi thu duoc
Đáp án:
$m_{muối} = 39,075(gam)$
Giải thích các bước giải:
$Fe + 2HCl → FeCl_2 + H_2$
$2Al + 6HCl → 2AlCl_3 + 3H_2$
$Zn + 2HCl → ZnCl_2 + H_2$
Theo phương trình trên , ta thấy :
$n_{HCl} = 2n_{H_2} = 2.\frac{6,72}{22,4} = 0,6(mol)$
Áp dụng ĐLBT khối lượng ,ta có :
$m_{\text{kim loại}} + m_{HCl} = m_{muối} + m_{H_2}$
$⇒ m_{muối} = 17,775 + 0,6.36,5 – 0,3.2 = 39,075(gam)$