How many grams of xenon and of fluorine are theoretically needed to make 1.000 g of XeF4?
2005-11-18 17:34:30 · 2 個解答 · 發問者 Frank 7 in 科學 ➔ 化學
好像算得太複雜了！又好像計算錯誤哦！
2005-11-19 04:56:25 · update #1
XeF4的分子量為131+19*4=207
Xe佔分子的131/207
而F佔分子的76/207
所以
Xe需要的重量為1*131/207=0.633g
F需要的重量為1*76/207=0.367g
2005-11-20 11:45:01 · answer #1 · answered by linichi.tw 6 · 0⤊ 0⤋
How many grams of xenon and of fluorine are theoretically needed to make 1.000 g of XeF4?
Molecular Weight of XeF4 = 131 + 4 x 19 = 207
1.000 g of XeF4 = 0.0048 mole of XeF4
2F2 + 1Xe --> XeF4
==> 0.0096 mole of F2 + 0.048 mole of Xe --> 0.048 mole of XeF4
so you need :
0.0096 x 38 = 0.367 g of F2
and
0.048 x 191 = 6.288 of Xe
2005-11-18 17:57:59 · answer #2 · answered by Anonymous · 0⤊ 0⤋