Calculation could somebody help me please?

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The discussion focuses on the calculations involved in chemical reactions, specifically the reaction of sodium metal with water and the neutralization of sodium hydroxide with sulfuric acid. The calculations include determining the number of moles of sodium, the molarity of sodium hydroxide produced, and the volume of hydrogen gas generated. Additionally, the molarity of sulfuric acid is calculated based on the neutralization reaction. The final results indicate a molarity of 0.167 M for sulfuric acid.

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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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