神马品牌网村长我的世界神奇宝贝:化学计算题
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将1.95克由硫酸钠和碳酸钠组成的混合物溶于水得溶液A,在A中加10毫升未知溶液的氯化钡溶液恰好完全反应,过滤的沉淀B和滤液C,在C中加足量硝酸银溶液,反映后又得沉淀5.74克,在沉淀B中加足量稀硫酸沉淀不消失,反而增加0.18克. 求氯化钡中溶质的物质的量浓度原混合物中碳酸钠的质量分数
(1)滤液C中的Cl-即为氯化钡溶液中Cl-
(Cl-)+(Ag+)==AgCl
1mol`````````````````143.5
x`````````````````````5.74
x=0.04mol
c=0.04mol/0.01L=4mol/L
(2)沉淀B中加稀硫酸:BaCO3转化为BaSO4,利用差量法
BaCO3+H2SO4==BaSO4+CO2+H2O 增重
197`````````````````233`````````````36
y````````````````````````````````````0.18
y=0.985g
BaCO3~Na2CO3
197````````106
0.985g```````z
z=0.53g
原混合物中碳酸钠的质量分数=(0.53g/1.95克)*100%=27.2%