勁人快來，會考化學mock題解答 超急。。。

The answer is (b).

Consider the reaction of 10 g of CaCO3 and excess HCl.
CaCO3 + 2HCl → CaCl2 + H2O + CO2
Mole ratio CaCO3 : CO2 = 1 : 1
No. of moles of 10 g CaCO3 = 10/100 = 0.1 mol
No. of moles of CO2 formed = 0.1 mol
Volume of 1 mol of a gas = 24 dm3
Hence, volume of CO2 = 0.1 x 24 = 2.4 dm3

(a)
Consider the reaction of 10 g of CaO and excess HCl.
CaO + 2HCl → CaCl2 + H2O
No CO2 is formed.
Therefore, if the impurity is CaO, volume of CO2 < 2.4 dm3
Actual volume of CO2 = 2.5 dm3
Hence, (a) is incorrect.

(b)
Consider the reaction of 10 g of NaHCO3 with excess HCl.
NaHCO3 + HCl → NaCl + H2O + CO2
Mole ratio NaHCO3 : CO2 = 1 : 1
No. of moles of 10 g NaHCO3 = 10/84 > 0.1 mol
No. of moles of CO2 formed > 0.1 mol
Volume of CO2 formed > 2.4 dm3
The same mass of NaHCO3 gives more CO2 than CaCO3.
If the impurity is NaHCO3, volume of CO2 > 2.4 dm3
Actual volume of CO2 = 2.5 dm3
Hence, (b) may be correct.

(c)
Consider the reaction of 10 g of KHCO3 with excess HCl.
KHCO3 + HCl → KCl + H2O + CO2
Mole ratio KHCO3 : CO2 = 1 : 1
No. of moles of 10 g KHCO3 = 10/100 = 0.1 mol
No. of moles of CO2 formed = 0.1 mol
Volume of CO2 formed = 2.4 dm3
Equal masses of KHCO3 and CaCO3 gives equal volumes of CO2.
If the impurity is KHCO3, volume of CO2 = 2.4 dm3
Actual volume of CO2 = 2.5 dm3
Hence, (c) is incorrect.

(d)
Consider the reaction of 10 g of FeCO3 with excess HCl.
FeCO3 + 2HCl → FeCl2 + H2O + CO2
Mole ratio FeCO3 : CO2 = 1 : 1
No. of moles of 10 g FeCO3 = 10/116 < 0.1 mol
No. of moles of CO2 formed < 0.1 mol
Equal masses of FeCO3 and CaCO3 gives less CO2 than CaCO3
If the impurity is NaHCO3, volume of CO2 < 2.4 dm3
Actual volume of CO2 = 2.5 dm3
Hence, (d) is incorrect.

我想若果可以用簡單的方法做找到答案,
就應該用了, 因考試時間不多,
在這一題裏, a一定不是答案, 因氧化物不能夠產生二氧化碳的.
而剩下b, c , d, 應選最少摩爾質量的.所以應選b了.
因在相同的質量裏, 最少摩爾質量就是最大的摩爾數了.