- #1
Janka
- 16
- 1
Hello people...could u help me pls...coz i am rly stuck in these thing like calculations...i know all right answers but i am not too sure how to get them...please if somebody of us has a few minutes could u help me with that?
Thank you very much.
1.,
When 0.25 g of sodium metal was added to 200 cm3 ( an excess ) of water , the following reaction occured.
Na(s) + H20 (l) --> NaOH ( aq ) + 1/2 H2 (g )
(i) calculate the number of moles of sodium taking part in the reaction.
Na = 23 g mol-1
0.25g / 23 g mol-1 = 0.011 mole
(ii) calculate the molarity of the sodium hydroxide solution which was
formed.
(iii) calculate the volume of hydrogen gas produced at r.t.p. (298 K and
100 kPa ). Assume that hydrogen is insoluble in water under these
conditions.
In another experiment 25.0 cm3 of 0.183 M sodium hydroxide were neutralised by 13.7 cm3 of sulphuric acid according to the following equation.
2NaOH (aq) + H2SO4 (aq) --> Na2SO4 (aq) + 2H2O ( l )
Calculate the molarity of the sulphuric acid.
Thank you very much.
1.,
When 0.25 g of sodium metal was added to 200 cm3 ( an excess ) of water , the following reaction occured.
Na(s) + H20 (l) --> NaOH ( aq ) + 1/2 H2 (g )
(i) calculate the number of moles of sodium taking part in the reaction.
Na = 23 g mol-1
0.25g / 23 g mol-1 = 0.011 mole
(ii) calculate the molarity of the sodium hydroxide solution which was
formed.
(iii) calculate the volume of hydrogen gas produced at r.t.p. (298 K and
100 kPa ). Assume that hydrogen is insoluble in water under these
conditions.
In another experiment 25.0 cm3 of 0.183 M sodium hydroxide were neutralised by 13.7 cm3 of sulphuric acid according to the following equation.
2NaOH (aq) + H2SO4 (aq) --> Na2SO4 (aq) + 2H2O ( l )
Calculate the molarity of the sulphuric acid.
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