Sản phẩm khi cho OA+bazơ

T

thupham22011998

Ta có: n $NaOH$=0,15 mol
Ta có tỉ lệ:

[TEX]1/2 < n CO_2/n NaOH <1 [/TEX]

-->Tạo 2 muối.

[TEX]2NaOH+CO_2-->Na_2CO_3+H_2O[/TEX]
0,15 ............0.075 ......0,075 mol
[TEX]CO_2+Na_2CO_3+H_2O-->2NaHCO_3[/TEX]
0,025........0,025..........................0,05 mol

Ta có: n $Na_2CO_3$=0,075-0,025=0,05 mol
-->m muối=0,05.106+0,05.84=9,5 g
 
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