2.60g of a metal X combine with 2.30g of oxygen to form an oxide
in which the oxidation number of X is+3.
What is the relative atomic mass of X? (Relative atomic mass: O=16.0)
A.11.6 B.34.7 C. 52.0 D. 104
其實我係有ans 不
過睇唔明 所有想請大家解一解d step係度做緊乜野
ANS
step 1) number of moles of O=1.2/16=0.075
step 2) In X2O3,mole ratio X:O=2:3
step 3) number of moles of X =(2/3)(0.075)=0.05
step 4) relative atomic mass of X=2.60/0.05=52
step 1個度 已經唔明
點解係number of moles of O而唔係number of moles of O2
step 3 點解要將個mole ratio乖番number of moles of O
唔該哂=]
CHEM計數題
2008-03-25 8:20 pm
回答 (2)
2008-03-25 9:15 pm
✔ 最佳答案
step1係計化合物X2O3中,氧的摩爾數,由摩爾數=質量/相對原子質量,number of moles of O=2.30/16X2=0.075
step2計化合物中X同O既比例,2:3
step3,設X的mole數為n,X與Omole數比為2/3=X/0.075,所以X的MOLE數為(2/3)0.075=0.05,所以就咁乘也可以
step4,根據摩爾數=質量/相對原子質量,變化為相對原子質量=質量/摩爾數=2.60/0.05=52
step1中,只要將2.3/2就係1.2了
2008-03-25 9:47 pm
你個答案係錯嘅!
4X + 3O2 --- > 2X2O3
no. of moles of X = 2.6 / M
no. of moles of O2 = 2.3 / 32 = 23 / 320
By moles ratio,
( 2.6 / M ) / ( 23 / 320 ) = 4 / 3
M = 27.1 ( 3 .s.f )
So r.a.m. of X = 27.1
4X + 3O2 --- > 2X2O3
no. of moles of X = 2.6 / M
no. of moles of O2 = 2.3 / 32 = 23 / 320
By moles ratio,
( 2.6 / M ) / ( 23 / 320 ) = 4 / 3
M = 27.1 ( 3 .s.f )
So r.a.m. of X = 27.1
收錄日期: 2021-04-23 19:57:58
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