chemical calculation...
The relative stomic mass of element X is 74.9. It forms an oxide containing 24.3% of oxygen by mass. What is the mole ratio of X to oxygen in the oxide?
( Relative atomic mass: O=16.0)
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- 匿名1 十年前最愛解答
Let 1:y be the mole ratio of X to oxygen in the oxide
[ (16*y) / (16*y + 74.9) ] x 100% = 24.3%
16y = 3.888 + 18.2007
y = 1.38 ( corr. to 3sig. fig.)
1 : 1.38 ~ 2 : 3
the mole ratio of X to oxygen in the oxide is 2:3
資料來源: Me ( 可能會計錯) - 1 十年前
let 100 g be the mass of the oxide
x=100-24.3=75.7g
no. of moles of x= 75.7/74.9=1.0107
no. of moles of o2= 24.3/32= 0.7594
relative no of moles= 4:3
i think the ratio should be 4:3
it may be wrong i am not sure
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