1...(from book p.98 - Q21.7)
2.65g of anhydrous sodium carbonate was dissolved in water and made up to 250.0cm3 solution. 25.0cm3 of this solution required 26.3cm3 of a hydrochloric acid solution for complets neutralization. Find the molarity of the hydrochloric acid.
2 ... (from book p.99 - Q21.8)
[a] Solution X contains 20.0g of sodium hydroxide per 250.0cm3 of solution. Calculate the molarity of the solution.
[b] Solution Y contains 18.0 g of acid (HnA) per 100.0 cm3 of solution. The molecular mass of the scid is 90. Calculate the molarity of the solution.
[c] 50.0 cm3 of solution are found to react completely with 25.0 cm3 of solution Y.
1* calculate th no. of moles of sodium hydroxide in 50.0cm3 of solution X.
2* calculate the no. of moles of HnA in 25.0 cm3 of solution Y.
3* how many moles of sodium hydroxide react with 1 moles of HnA?
4* what is the value of n in HnA?
5* write an equation for the reaction which takes place between solutions X and Y.
Chemistry -- Question x 2
2008-04-15 5:21 am
回答 (2)
2008-04-15 7:38 am
✔ 最佳答案
1. Molar mass of Na2CO3 = 23x2 + 12 + 16x3 = 106 g mol-1
No. of moles of Na2CO3 in titration = (2.65/106) x (25/250) = 0.0025 mol
Na2CO3 + 2HCl → 2NaCl + H2O + CO2
No. of moles of HCl in titration = 0.0025 x 2 = 0.005 mol
Volume of HCl solution = 26.3 cm3 = 0.0263 dm3
Molarity of HCl = mol/V = 0.005/0.0263 = 0.19 M
==========
2 . [a]
Molar mass of NaOH = 23 + 16 + 1 = 40 g mol-1
No. of moles of NaOH = mass/(molar mass) = 20/40 = 0.5 mol
Volume of NaOH solution = 250 cm3 = 0.25 dm3
Molarity of NaOH = mol/V = 0.5/0.25 = 2 M
==========
2 . [b]
No. of moles of HnA = mass/(molar mass) = 18/90 = 0.2 mol
Volume of HnA solution = 100 cm3 = 0.1 dm3
Molarity of HnA = mol/V = 0.2/0.1 = 2 M
==========
2 .[c] 1*
No. of moles of NaOH = MV = 0.2 x (50/1000) = 0.01 mol
2 .[c] 2*
No. of moles of HnA = MV = 0.2 x (25/1000) = 0.005 mol
2 .[c] 3*
No. of moles of NaOH react with 1 mole of HnA
= 0.01/0.005
= 2
2 .[c] 4*
2 mol of NaOH reacts with 2 mol of ionizablbe H atom (H+).
2 mol of NaOH reacts with 1 mol of HnA.
Hence, n = 2
2 .[c] 5*
H2A + 2NaOH → Na2A + 2H2O
2008-04-15 7:49 am
1...
首先寫equation: Na2CO3 2HCl->2NaCl CO2 H2O
no. of moles of Na2CO3=2.65/(23x2 12 16x3) [vol/relative atomic mass]
=0.025mol
no. of moles used=0.025x25/250=0.0025mol [25cm3 out of 250cm3 used]
From the equation,2moles of HCl react with 1mole of Na2CO3.
Thus no. of moles of HCl=0.0025x2=0.005
Molarity of HCl=0.005/(26.3/1000) [no. of moles/vol(in dm3)]
=0.190M
2...
[a]Molarity={20/(23 16 1)}/(250/1000) [no. of moles of NaOH/vol(in dm3)]
=2M
[b]Molarity=(18/90)/(100/1000)
=2M
[c]1* no. of moles={20/(23 16 1)}x50/250
=0.1mol
2* (18/90)x25/100=0.05mol
3* 0.1/0.05=2mol
4* n=2 [睇eqution...左邊要2個Na,,所以右邊NanA->n=2]
5* 2NaOH H2A->Na2A 2H2O
希望幫到你
首先寫equation: Na2CO3 2HCl->2NaCl CO2 H2O
no. of moles of Na2CO3=2.65/(23x2 12 16x3) [vol/relative atomic mass]
=0.025mol
no. of moles used=0.025x25/250=0.0025mol [25cm3 out of 250cm3 used]
From the equation,2moles of HCl react with 1mole of Na2CO3.
Thus no. of moles of HCl=0.0025x2=0.005
Molarity of HCl=0.005/(26.3/1000) [no. of moles/vol(in dm3)]
=0.190M
2...
[a]Molarity={20/(23 16 1)}/(250/1000) [no. of moles of NaOH/vol(in dm3)]
=2M
[b]Molarity=(18/90)/(100/1000)
=2M
[c]1* no. of moles={20/(23 16 1)}x50/250
=0.1mol
2* (18/90)x25/100=0.05mol
3* 0.1/0.05=2mol
4* n=2 [睇eqution...左邊要2個Na,,所以右邊NanA->n=2]
5* 2NaOH H2A->Na2A 2H2O
希望幫到你
參考: 自己...
收錄日期: 2021-04-19 21:37:31
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