Calculation could somebody help me please?

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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 occurred.

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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Concentration = moles/volume

Molarity = Concentration

If you need further help ask away...
 
okay... i'll explain it to you...

(1) Ar Na: 23 therefore, 0.25 g Na contains (0.25/23) mol = 0.011 mol

(2) number of moles NaOH formed:

1 mol Na gives 1 mol NaOH
therefore, 0.011 mol Na will give 0.011 NaOH

200 cm3 (approx.) contains 0.011 mol NaOH
100 cm3 will contain (0.011*100/200) = 0.005 mol (approx.)

this is the molarity, 0.005 mol/dm3

(3) NUMBER OF MOLE OF h2 PRODUCED: (0.5*0.011) since 1 mol Na produces 0.5 mol H2 = 0.005 mol
at r.t.p. 1 mol of gas occupies 24 dm3
therefore 0.005 mol H2 will occupy (0.005*24) = 0.130 dm3 = 130 cm3


number of mol NaOH in 25 cm3 = (0.183*25/1000) since 1000 cm3 contains 0.183 mol NaOH = 0.005 mol

therefore, since 2 mol NaOH reacts with 1 mol H2SO4, the number of mol of H2SO4 which reacted is (0.005/2) = 0.002 mol

13.7 cm3 contains 0.002 mol H2SO4
1000cm3 will therefore contain (0.002*1000/13.7) = 0.167 mol H2SO4

0.167 M
 

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