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cyw1226 2013-8-22 12:45 AM

relative atomic mass

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chungkin81 2013-8-22 09:02 AM

The formula of a metal carbonate is X2CO3. 50.0 cm3 of a solution containing 0.53 g of the carbonate requires 25.00 cm3 of 0.2 M sulphuric acid for complete neutralization. What is the relative atomic mass of metal X?

Solution:
X2CO3 (aq) + H2SO4 (aq) --> X2SO4 (aq) + CO2 (g) + H2O (l)
No. of moles of H2SO4 = 0.2 x 25.00 / 1000 = 5 x 10^-3 mol
No. of moles of X2CO3 = 5 x 10^-3 = 5 x 10^-3 mol

In 1 mole of X2CO3, it contains 1 mole of CO3^2-
No. of moles of CO3^2- = 5 x 10^-3 mol
Mass of CO3^2- = 5 x 10^-3 x (12 + 16 x 3) = 0.30 g
Mass of X = 0.53 - 0.30 = 0.23 g

In 1 mole of X2CO3, it contains 2 moles of X
No. of moles of X = 5 x 10^-3 x 2 = 0.01 mol
Let the relative atomic mass of X be M
0.23 / M = 0.01
M = 23
The relative atomic mass of X is 23.
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查看完整版本: relative atomic mass