0.5 M 60cm^3 Sulphuric acid is mixed with 0.2M 200 cm^3 Sodium Hydroxide solution in a beaker.

a. Write an equation for the above reaction.

b. State an evidence for the reaction taken place.

c. Calculate the no. of moles of sulphuric acid & sodium hydroxide solution used respectively.

d. After the reaction, one of the productis in excess.

Show by calculation, which reactant is in excess.


e.(i) Write an equation for this reaction.

(ii) What do you observe in the reaction.

(iii) Calculate the no. of moles of aluminium needed for complete reaction with the excess reactant in (d).

(iv) Calculate the mass of aluminium needed.

2 Answers

  • 1 decade ago
    Favorite Answer

    a. H2SO4+2NaOH->2H2O+Na2SO4

    b. Temperature of the reaction mixture rise, because the reaction is exothermic.

    c. H2SO4: 60/1000*0.5=0.03mol

    NaOH: 200/100*0.2=0.04mol

    d. According to equation, no. of mol of H2SO4: no. of mol of NaOH is 1:2. 0.04mol of

    NaOH requires 0.04/2mol of H2SO4= 0.02Mol. There is still 0.03-0.02=0.01 mol of H2SO4 remain in the mixture after reaction. H2SO4 is in excess. (我唔知點解佢問"PRODUCT IN EXCESS",應該問 " REACTANT IN EXCESS".

    問題d.之後係咪有一段文字話用Al O黎REACT 埋O的ACID?

    e.(i) 2Al+3H2SO4->Al2(SO4)3+3H2

    (ii) Bubble given out, temperature rise.

    (iii)no. of mol of H2SO4: no of mol of Al=3:2, no. of mol of Al required:0.01*2/3=0.0067mol(2sig.fig.)


  • 1 decade ago

    a) H2SO4+2NaOH --> Na2SO4+ 2H2O

    b) Heat is released. Since this reaction is a reaction of neutralization.

    c) no.of mole of H2SO4=(0.5)(60/1000)=0.03

    no. of mole of NaOH=(0.2)(200/1000)=0.04

    d) NaOH is in excess, since its no. of mole is larger than the no. of mole of H2SO4

    *Molarity=no. of mole of solute/ vol. of solute

    **unit of the vol. should be dm^3

    I'm not quite sure for the ans. of (d)

    Source(s): me
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