Metal M forms a chloride of formula MCl2. 6.30g of M are added to 50.0cm3 of
5.00M HCl. After the reaction has completed, the solution is diluted to 250cm3.
25.0cm3 of the diluted solution require 15.6cm3 of a 0.0800 M Na2CO3 for
complete neutralisation
a) Write appropriate equations in this experiment.
b) Find the relative atomic mass of M.
*Please show steps in detail. Thanks:)
Chem titration calculation
2013-11-14 4:49 am
回答 (3)
2013-11-14 6:14 am
✔ 最佳答案
Na₂CO₃ + 2HCl --> 2NaCl + CO₂ + H₂Ono. of mole of Na₂CO₃ = 0.08x(15.6/1000) = 0.001248
no. of mole of HCl remains in 25 cm³ diluted solution = 0.001248x2 = 0.002496
no. of mole of HCl remains in 25 cm³ diluted solution = 0.002496x(250/25) = 0.02496
no. of mole of HCl in 50 cm³ 5.00M HCl = 5.00x(50/1000) = 0.25
M + 2HCl --> MCl₂ + H₂
no. of mole of HCl reacted with M = 0.25 - 0.02496 = 0.22504
no. of mole of M = 0.22504/2 = 0.11252
atomic mass of M = 6.30/0.11252 = 56.0 g / mole
2013-11-14 11:28 pm
Relative atomic mass has no unit.
2013-11-14 5:32 am
a) M + 2HCl -> MCl2 + H2
b) no of mol of Na2CO3 used = 0.08 * 15.6 / 1000
= 0.00125
2HCl + Na2CO3 -> 2NaCl + H2CO3
no of mol of HCl remained = 0.00125 * 2 / 10
= 0.00025
the relative atomic mass of M = 6.3 / 1000 / 0.00025
= 25.2
2013-11-13 21:34:52 補充：
b) molarity = no of mol / volume in dm^3
& molarity = mass / molar mass
2013-11-14 12:25:49 補充：
係喎, 唔記得0.00025 係remain HCl
我個錯了
b) no of mol of Na2CO3 used = 0.08 * 15.6 / 1000
= 0.00125
2HCl + Na2CO3 -> 2NaCl + H2CO3
no of mol of HCl remained = 0.00125 * 2 / 10
= 0.00025
the relative atomic mass of M = 6.3 / 1000 / 0.00025
= 25.2
2013-11-13 21:34:52 補充：
b) molarity = no of mol / volume in dm^3
& molarity = mass / molar mass
2013-11-14 12:25:49 補充：
係喎, 唔記得0.00025 係remain HCl
我個錯了
收錄日期: 2021-04-13 19:48:41
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