题干

 胃酸主要成分是盐酸,正常人胃液每日需要盐酸溶质质量约为6.5g~7.3g.某患者每日分泌盐酸溶质质量为9.0g,为治疗胃酸过多需服用图所示抑酸剂,发生反应:Mg(OH)2+2HCl=MgCl2+2H2O.请计算并回答:按说明书用法用量,每天最多可中和胃液中盐酸溶质质量是多少克?理论上能否使该患者胃液中盐酸溶质质量恢复到正常范围?(写出计算过程,结果精确到0.001)

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答案(点此获取答案解析)

解:设每天最多可消耗胃液中盐酸溶质质量为m,

Mg(OH)2+2HCl=MgCl2+2H2O

58 2×36.5

0.145×2×3 m

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