F.4 Chemistry "Reacting mass" 問題(3)

e家答唔知有冇遲到
其實呢題寫唔寫方程都OK 只要概念通同善用表格就計到
咁開始了
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no. of mole of water=1.44/18=0.08mol
since 1 mole of water contain 1mole of oxygen atoms
so no. of mole of oxygen=0.08mole
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之後用表
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atom________________X__________________O
MASS______13.7-0.08X16=12.42g ________1.28g
no of mole_________0.06mol_____________ 0.08mol
ratio_________________1 _______ : _______1.333(0.08/0.06)
____________________3_________:_________4
there for , the empirical formula of the oxideis X3O4
唔 又係錯解????

2007-01-19 23:10:54 補充：
提醒你一樣野雖然你可以用週期表搵返果隻係咩METAL 但係其實計MOLE數題目既黎就免啦因為好多時同我地既期望結果唔同例如S同CL咁平時應該係SCL2但係佢都可以用數字搞到佢變S2CL2所以 小心啦

你係男定女呀, 唔好一次過問咁多



number of mole of 13.7g XOy = 13.7 / ( 207 + 16y )

number of mole of 1.44g of H2O = 1.44 / ( 2 + 16 ) = 0.08 mole

Write out the equation : XOy + y H2 -----> X + y H2O


Hence the number of mole of XOy : H2O = 1 : y

==> 13.7 / ( 207 + 16y ) : 0.08 = 1 : y

Solve for y

==> 13.7 y = 0.08 ( 207 + 16y )

==> y = 4/3



Hence this "should be" ... X3O4 , but actually is the mixing of XO and XO2