Hỗn hợp có khối lượng 7,6g gồm Mg và MgO được chia làm 2 phần bằng nhau phần 1 t/d vs dd HCl dư thu đc 1,68l khí(đktc) phần 2 t/d vừa hết vs m gam dd H2SO4 24,5%

1 câu trả lời

Đáp án:

 $m_{dung dịch H_2SO_4}= 88g$

Giải thích các bước giải:

Phần 1:

Gọi $n_{Mg}= a mol$, $n_{MgO}= b mol$

 $n_{H_2}=\frac{1,68}{22,4}=0,075 mol$
phương trình phản ứng

$Mg + 2HCl \to MgCl _2 + H_2$
$MgO + 2HCl \to MgCl_2 + H_2O$
$n_{Mg}=n_{H_2}= 0,075 mol$
$n_{MgO}=\frac{7,6-0,075.24}{40}=0,145 mol$
Phần 2:

$Mg + H_2SO_4 \to MgSO_4 + H_2$
$MgO + H_2SO_4 \to MgSO_4 + H_2O$
$n_{H_2SO_4}= 0,075+0,145= 0,22 mol$
$m_{dung dịch H_2SO_4}=\frac{0,22.98.100}{24,5}=88g$

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