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    (Original post by Habeeb H Patel)
    When silicon tetrachloride is added to water, the following reaction occurs: SiCl4+2H2O=SiO2+4HCL 1.2g of impure silicon tetrachloride was dissolved in excess water and the resulting solution was made up to 250cm^3. A 25^3 portion was the titrated against 0.10 moldm-3 sodium hydroxide and 18.7cm^3 of the alkali were required. What was the percentage purity of the silicon tetrachloride. I know how to work it out, but dont understand why NaOH has the same number of moles as HCl....... or is that just an assumption because you dont know the ratio between them
    So you are right in finding the number of moles of NaOH in 18.7cm^3 of the 0.1moldm^-3 solution. The number of moles of NaOH in this reaction is the same number of moles of HCl because the molar ratio in this neutralisation reaction is 1:1

    You can tell the molar ratio of any reaction by writing the equation for it. So in this case you have a neutralisation reaction:
    NaOH + HCl --> NaCl + H2O

    Because everything is balanced, the molar ratio of the reaction is 1:1 --> 1:1, i.e. 1 mole of sodium hydroxide with 1 mole of hydrochloric acid always gives 1 mole of water and 1 mole of sodium chloride.

    Then once you have the moles of NaOH, and therefore HCl, you can work backwards to answer the question.
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    No I'm not clever enough
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    (Original post by Caecilius_)
    So you are right in finding the number of moles of NaOH in 18.7cm^3 of the 0.1moldm^-3 solution. The number of moles of NaOH in this reaction is the same number of moles of HCl because the molar ratio in this neutralisation reaction is 1:1

    You can tell the molar ratio of any reaction by writing the equation for it. So in this case you have a neutralisation reaction:
    NaOH + HCl --> NaCl + H2O

    Because everything is balanced, the molar ratio of the reaction is 1:1 --> 1:1, i.e. 1 mole of sodium hydroxide with 1 mole of hydrochloric acid always gives 1 mole of water and 1 mole of sodium chloride.

    Then once you have the moles of NaOH, and therefore HCl, you can work backwards to answer the question.


    Thank you so much, it makes sense now and taking your time to answer. Much appreciated
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    (Original post by Caecilius_)
    So you are right in finding the number of moles of NaOH in 18.7cm^3 of the 0.1moldm^-3 solution. The number of moles of NaOH in this reaction is the same number of moles of HCl because the molar ratio in this neutralisation reaction is 1:1

    You can tell the molar ratio of any reaction by writing the equation for it. So in this case you have a neutralisation reaction:
    NaOH + HCl --> NaCl + H2O

    Because everything is balanced, the molar ratio of the reaction is 1:1 --> 1:1, i.e. 1 mole of sodium hydroxide with 1 mole of hydrochloric acid always gives 1 mole of water and 1 mole of sodium chloride.

    Then once you have the moles of NaOH, and therefore HCl, you can work backwards to answer the question.


    Now lets say hypothetically, the ratio between NaOH and HCL was 2:1, the how would you go about to answer the question
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    (Original post by Habeeb H Patel)
    Now lets say hypothetically, the ratio between NaOH and HCL was 2:1, the how would you go about to answer the question
    If that were the case, after working out the moles of NaOH you would half that number to work out the number of moles of HCl because the ratio is 2:1, so for every 2 moles of NaOH there is 1 mole of HCl.

    In this case the ratio is 1:1 so if you do the maths there are 0.00187 moles of HCl in the 25cm^3 solution
 
 
 
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