Dùng `V` ml dung dịch Ba(OH)2 0,5M hấp thụ hoàn toàn 2,24 lít khí SO2(đktc ) . Sau phản ứng thu được muối BaSO3 không tan . Giá trị bằng số của `V` là bao nhiêu ?

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Bạn tham khảo!
$\text{Ta có PTHH sau:}$

$\text{Ba(OH)2+SO2$\rightarrow$BaSO3↓+H2O}$

$\text{Có nSO2=$\frac{V}{22,4}$=$\frac{2,24}{22,4}$=0,1 mol}$

$\text{Có nBa(OH)2=nSO2=0,1 mol}$

$\rightarrow$ $\text{V Ba(OH)2=$\frac{n}{CM}$=$\frac{0,1}{0,5}$=0,2 lít}$

$\text{Đổi 0,2 lít thành 200ml}$

$\text{Vậy giá trị của V là 200}$

#TRANPHAMGIABAO

 

Ba(OH)2 + SO2 -> BaSO3 + H2O

 nSO2 = 2,24/22,4 = 0,1 (mol)

=> nBa(OH)2 = 0,1 (mol) (do Ba(OH)2 và SO2 tỉ lệ 1:1)

=> V = n/CM = 0,1/0,5 = 0,2 (lít) = 200ml

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