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化學的摩爾數

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< async src="//pagead2.googlesyndication.com/pagead/js/adsbygoogle.js"> 發問:

有60g硝酸鋁...Al(NO3)3 ........兩個3字都係右下角.求: a)陽離子的摩爾數 b)陰離子的摩爾數 c)陽離子數目 d)陰離子數目 相對原子質量:N=14;O=16:Al=27 唔該要有解釋同步驟 THX

最佳解答:

Relative formula mass of N = 14.0, O = 16.0 and Al = 27.0 Molar mass of Al(NO3)3 = 27.0 + (14.0 + 16.0 X 3) X 3 = 213.0 g/mol In 60 g of Al(NO3)3, no. of mole of Al(NO3)3 = mass / molar mass = 60 / 213 = 20 / 71 mol a. The no. of mole of the cations, Al3+ = 20 / 71 mol = 0.282 mol b. The no. of mole of the anions, NO3- = 20 / 71 X 3 = 60 / 71 mol = 0.845 mol For the following calculations, they are based on the Avodagro's number. i.e., there are 6.02 X 1023 particles in one mole of substances. c. no. of cations = no. of mole X Avodagro's no. = 20 / 71 X 6.02 X 1023 = 1.70 X 1023 d. no. of anions = no. of mole X Avodagro's no. = 60 / 71 X 6.02 X 1023 = 5.09 X 1023

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