CE CHEM Volumetric analyze

2008-03-23 3:37 am

b part 唔識做,請幫忙,謝謝

1. 3.60g of a solid mixture of sodium carbonate-10-water and sodium carbonate
Required 23.0cm cube of 1.50M hydrochloric acid for complete reaction
A(i) WRITE chemical equations of the reaction between
(1) sodium carbonate and hydrochloric acid, and
(2) sodium hydrogencarbonate and hydrochloric acid
B Calculate the percentage by mass of each substance in the mixture
(R.A.M : H =1 , C=12 ,O=16,Na=23)
Ans : (B) 47.3% sodium carbonate-10-water
52.3% sodium hydrogencarbonate

回答 (1)

用戶7369

2008-03-23 5:16 am

✔ 最佳答案

b. let x be the mass of hydrated sodium carbonate
(3.6-x ) is the mass of sodium bicarbonate.
the no. of mole of sodium carbonate used=x/286
the no. of mole of sodium bicarbonate used=(3.6-x)/84
the no. of mole of acid used=23/1000 x 1.5
therefore, 2(x/286) + (3.6-x)/84 = 23/1000 x 1.5
after solving, x=1.7
the %of sodium carbonate by mass =1.7/3.6x100=47.2%
the % of sodium bicarbonate by mass=1.9/3.6x100%=52.7%

👍 0 👎 0