chem 計數問題

X(g) + 3Y(g) → 2Z

No. of moles of X = 40/24000 = 0.00167 mol
No. of moles of Y = 60/24000 = 0.0025 mol

Mole ratio X : Y = 1 : 3
When Y is completely reacted,
no. of moles of X needed = 0.0025 x (1/3) = 0.00083 mol
Therefore, X is in excess and Y is the limiting reactant.
No. of moles of X unreacted = 0.00167-0.00083 = 0.00084 mol
Volume of X unreacted = 0.00084 x 24000 = 20 cm3

Mole ratio Y : Z = 3 : 2
No. of moles of Y reacted = 0.0025 mol
No. of moles of Z formed = 0.0025 x (2/3) = 0.00167 mol
Volume of Z formed = 0.00167 x 24000 = 40 cm3

Volume of the resultant gaseous mixture
= (Volume of X unreacted) + (Volume of Z formed)
= 20 + 40
= 60 cm3

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Alternative method (a better method)
According to Avogadro’s law, under constant pressure and temperature, the volume ratio of gases is equal to the mole ratio.

X(g) + 3Y(g) → 2Z

Volume ratio X : Y = 1 : 3
When 60 cm3 of Y is completely reacted,
volume of X needed = 60 x (1/3) = 20 cm3
But there is 40 cm3 of X.
Hence, X is in excess, and Y is the limiting reactant.
Volume of X unreacted = 40 - 20 = 20 cm3

Volume ratio Y : Z = 3 : 2
Volume of Y reacted = 60 cm3
Volume of Z reacted = 60 x (2/3) = 40 cm3

Volume of the resultant gaseous mixture
= (Volume of X unreacted) + (Volume of Z formed)
= 20 + 40
= 60 cm3

發問者和002答案都係錯在唔識英文，唔知乜野叫做 "resultant gaseous mixture"。

其實意思係問你結果既氣體混合物既體積，淨係計 Z 生成 40 cm3 冇乜用，要加埋剩低 20 cm3 先得架。答案真係 B 60 cm3，唔好死話係 A 先得架。唔識解題目要問下識既人，唔好亂教人，好易累死人架。叫人check數之餘都要自己學好D英文先得架。

我去投票之前，答啱既001答案同答錯既002答案取然同票，真係 ....

你無計錯~X亦係excess~個答案真係A

不過想提一提你~因為le個reaction入面reactants同product全部都係gas~所以mole ratio = volume ratio~原因係喺S.T.P & R.T.P底下~1 mole ge任何gas都有特定volume(22.4L & 24.5L respectively)~不過記得mole ratio = volume ratio le個情況之適用於reactants & product(s)全部都係gas ge時候先用得

所以你條數可以咁計法↓↓↓(你戈個計法都岩ga)
mole ratio X: Y: Z = 1: 3: 2
60cm^3 fo Y produces 40cm^3 of Z (直接60cm^3 / 3 * 2)
40cm^3 of X produces 80cm^3 of Z (直接40cm^3 * 2)
so Y is the limiting reagent, X is the excess reagent
so volume of Z = 40cm^3

↑↑↑記得la~test & exam戈陣可以省返d時間做其他題or覆卷a^^