請幫我解答一下化學題

no. of mole of HCl produce per minutes
= 0.5 * 20 = 10 mol
2 HCl + Ca(OH)2 --> CaCl2 + 2 H20
mole ratio = no. of mole of HCl : no. of mole of Ca(OH)2
= 2 : 1
no. of mole of Ca(OH)2 used = 0.5 x no. of mole of HCl produced
= 0.5 x 10
= 5 mol
mass of Ca(OH)2 used
= no of mole of Ca(OH)2 x molar mass of Ca(OH)2
= 5 (40+(16+1)x2)
= 370g

每分鐘排放氫氯酸的mole 數= 20x 0.5 = 10 moles

1mol 氫氯酸 (HCl) react with 0.5 mol 氫氧化鈣( Ca(OH)2 )
2HCl + Ca(OH)2 --> CaCl2 + 2 H2O

因此, 10 mol 氫氯酸需要 5 mol 氫氧化鈣來中和
1 mol 的氫氧化鈣 重: 40.1+ (16+1)x2 = 74.1g

因此,每分鐘需要 5x 74.1 = 370.5 g 的氫氧化鈣來中和

2HCl + Ca(OH)2 → CaCl2 + H2O + H2

氫氯酸的犘爾數量 = 0.5 x 20/1000

= 0.01 mol

氫氯酸的犘爾數量 比 氫氧化鈣的犘爾數量 = 2 : 1

∴氫氧化鈣的犘爾數量 = 0.01 / 2

= 0.005 mol

氫氧化鈣的質量 = 犘爾數 x 犘爾質量

= 0.005 x ( 40.1 + 2 x 16 + 2 x 1 )

= 0.3705 g

∴每分鐘需用多少熟石灰是 0.3705 g