问题 计算题

称取63.2 g KMnO4固体,先加热分解制取一定量的O2,再将剩余固体和足量36.5%的浓盐酸()混合加热制取Cl2。设收集到的O2mol,收集到的Cl2mol,最后Mn元素全部以Mn2+的形式存在于溶液中。

(1)如果配制上述500 mL浓盐酸,则需标准状况下HCl气体的体积为            L。

(2)当=0.16时,收集到Cl2的物质的量为                        

答案

(1)134.4(3分)  (2)0.68mol(4分)

考查物质的量的有关计算。

(1)根据可知,盐酸的浓度是,所以氯化氢的物质的量是12mol/L×0.5L=6mol,则在标准状况下的体积是6mol×22.4L/mol=134.4L。

(2)高锰酸钾的物质的量是63.2g÷158g/mol=0.4mol。根据电子的得失守恒可知,n(O2)×4+n(Cl2)×2=n(KMnO4)×5,即0.16mol×4+2n(Cl2)=0.4mol×5,解得n(Cl2)=0.68mol。

阅读理解

阅读表达

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名词解释