CHEM 計算

1) 所用的 HCl 的摩爾數 = 0.5 × (34/1000) = 0.017 mol
由方程: 2HCl + CaCO3 → CaCl2 + CO2 + H2O
一摩爾的碳酸鈣需要兩摩爾的 HCl 完全反應.
所以碳酸鈣樣本中所含碳酸鈣的摩爾數為 0.017/2 = 0.0085 mol
即碳酸鈣樣本中所含碳酸鈣的質量為: 0.0085 × 100 = 0.85 g
所以樣本內碳酸鈣純度百分比 = 0.85/1.28 × 100% = 66.4%
(2) 由方程: HNO3 +NaOH → NaNO3 + H2O
一摩爾的 NaOH 需要一摩爾的 HCl 完全反應.
所以混合物中過量的酸的摩爾數為: 0.4 × (20/1000) = 0.008 mol
而當初的硝酸摩爾數為: 0.5 × (50/1000) = 0.025 mol
所以與碳酸鈣反應的硝酸摩爾數為 0.025 - 0.008 = 0.017 mol
由方程: 2HNO3 +CaCO3 → Ca(NO3)2 + H2O + CO2
一摩爾的碳酸鈣需要兩摩爾的 HNO3 完全反應.
所以碳酸鈣樣本中所含碳酸鈣的摩爾數為 0.017/2 = 0.0085 mol
即碳酸鈣樣本中所含碳酸鈣的質量為: 0.0085 × 100 = 0.85 g
所以樣本內碳酸鈣純度百分比 = 0.85/3 × 100% = 28.3 %

2007-03-29 23:15:19 補充：
HNO3 是酸, NaOH 是鹼, 所以它們相加時會中和.第二題的第二步用 NaOH 目的為求出尚未反應的 HNO3 的摩爾數, 從而求出當初反應了的 HNO3 的摩爾數.

1.
CaCO3 + 2HCl ---> CaCl2 + H2O + CO2
Mole ratio of CaCO3:HCl = 1:2
No. of moles of CaCO3 = (34/1000)(0.5)/2 = 0.0085 mol
Mass of CaCO3 = 0.0085*(40.1+12.0+16.0*3) = 0.85085 g
Percentage purity = 0.85085/1.28*100% = 66.5%

2.
CaCO3 + 2HNO3 ---> Ca(NO3)2 + H2O + CO2---(1)
NaOH + HNO3 ---> NaNO3 + H2O---(2)

From equation (2), mole ratio of NaOH:HNO3=1:1
No. of moles of HNO3 (excess material) = (20/1000)(0.4) = 0.008 mol
No. of moles of HNO3 reacted in equation (1) = (50/1000)(0.5)-0.008 = 0.017 mol

From equation (1), mole ratio of CaCO3:HNO3 = 1:2
No. of moles of CaCO3 = 0.017/2 = 0.0085 mol
From question (1), percentage purity = 66.5%

2007-03-29 21:48:33 補充：
2. (amendment starting from line 6, not counting skipped lines)From equation (1), mole ratio of CaCO3:HNO3 = 1:2No. of moles of CaCO3 = 0.017/2 = 0.0085 molMass of CaCO3 = 0.0085*(40.1 12.0 16.0*3) = 0.85085 gPercentage purity = 0.85085/3*100% = 28.4%