摘要:解:设碳酸钠的质量为x .碳酸氢钠的质量为 y .所需盐酸溶液的物质的量浓度为C HCl . 2NaHCO3 ~~~~~~~~~~~CO2~~~~~~~~~~~CaCO3 186 44 100 y 2 g y =186×2 g/100 = 3.36 g x = 4.42 g - y = 4.42 g - 3.36 g = 1.06 g (2)2NaHCO3 ~~~~~~~~~~~Na2CO3~~~~~~~~~~~2HCl 286g 2mol 3.36g n1 Na2CO3~~~~~~~~~~~2HCl 106g 2mol

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