hoá khó

L

lovelycat1994

S

socviolet

[TEX]n_{C_3H_{12}O_3N_2=0,015; n_{NaOH}=0,04[/TEX]
(CH3NH3)2CO3 + 2NaOH ---> Na2CO3 + 2CH3NH2 + 2H2O
0,015---------------->0,03-------->0,015mol
=> NaOH dư 0,01mol => m=0,015.106+0,01.40=1,99g
 
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