【題目】已知硅能與堿反應(yīng)生成氫氣,將鎂、鋁、硅的混合物分為等質(zhì)量的兩份,一份跟足量的NaOH溶液反應(yīng),另一份跟足量的鹽酸反應(yīng),最終產(chǎn)生的H2一樣多,則鎂與硅的物質(zhì)的量之比為多少?

【答案】2∶1

【解析】Mg不和NaOH反應(yīng),所以

2Al+2NaOH+2H2O===2NaAlO2+3H2↑,

Si+2NaOH+H2O===Na2SiO3+2H2。

Si不和HCl反應(yīng),所以

2Al+6HCl===2AlCl3+3H2↑,

Mg+2HCl===MgCl2+H2↑。

因?yàn)楫a(chǎn)生的H2一樣多,等量的Al分別產(chǎn)生的H2是1∶1,

所以Si和Mg產(chǎn)生的氫氣就該是1∶1,

所以由方程式得n(Mg)∶n(Si)=2∶1。

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