合金x只含有鋅和銅金屬。若把1.0g合金x放入大量稀氫氯酸中,會產生270cm^3無色氣體。根據以上資料,合金x中含有銅的質量百分比是多少??
(相對原子質量:Zn=65.4 , Cu=63.5;在常溫常壓下,氣體的摩爾體積=24dm^3)
化學摩爾計算題
2007-04-30 9:59 pm
回答 (2)
2007-04-30 10:10 pm
✔ 最佳答案
首先, 利用鋅會和稀氫氯酸反應釋出氫氣而銅不會, 和以下方程式:Zn(s) + 2HCl(aq) → ZnCl2(aq) + H2(g)
便可得知一摩爾的鋅金屬可釋出一摩爾的氫氣.
如此, 由題目可知釋出的氫氣摩爾數 = 270 / 24000 = 0.01125 mol
因此, 在合金 x 中鋅的摩爾數亦為 0.01125 mol, 即 0.01125 × 65.4 = 0.07358 g
最後在合金 x 中銅的質量為 1 - 0.07358 = 0.2642 g
所以合金 x 中含有銅的質量百分比為 26.42%.
參考: My chemical knowledge
2007-04-30 10:12 pm
only Zn reacts with acid HCl(aq)
Zn(s) + 2H+(aq) --> H2(g) + Zn2+(aq)
the colurless gas is H2
the stoichemistric ratio of Zn to H2 is 1:1
no. of mole of H2 evolved=270*10^-3/24=0.01125mol
mass of Zn in the alloy X=0.01125*65.4=0.7358g
percentage by mass of Cuin the alloy X=(1.0-0.7358)/1.0=26.43%
Zn(s) + 2H+(aq) --> H2(g) + Zn2+(aq)
the colurless gas is H2
the stoichemistric ratio of Zn to H2 is 1:1
no. of mole of H2 evolved=270*10^-3/24=0.01125mol
mass of Zn in the alloy X=0.01125*65.4=0.7358g
percentage by mass of Cuin the alloy X=(1.0-0.7358)/1.0=26.43%
參考: myself
收錄日期: 2021-04-22 00:31:30
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20070430000051KK01691