Chemistry

posted by Sally

Yesterday we combined Hydrochloric Acid HCl with Sodium Hydroxide NaOH in a violent reaction that resulted in water H2O and common table salt NaCl.

How many grams of hydrochloric acid should we use so we have exactly enough to react with 40g of sodium hydroxide? You'll need to write a chemical reaction, balance it, and then perform your calculation.

also how many grams of salt NaCl will be produced and how many grams of water H2O be produced?

THANK YOU <3

  1. Barry

    Dear Sally,
    as I have replied you before, I am not sure whether you have checked
    HCL(aq) + NAOH(aq)---- NACL(s)+ H20(l)

    Number of moles in the NAOH(aq)=
    40/(23+16+1)
    =1
    as 1 mole of HCL will react with 1 mole of the Naoh solution
    number of mole of HCL required= 1

    Still remember that there is a equation
    mol= g/molar mass
    Let the number of grams of HCL needed be g grams
    the number of grams of HCL required=
    1= g/(1+35.5)
    g=36.5
    Therefore, 36.5 grams of HCL is required

    For the grams of Nacl produced
    1 mole of Hcl will produce 1 mole of nacl
    Let the number of grams of salt produced be g grams
    1x(23+35.5)
    =55.5
    Therefore 55.5 grams of nacl is produced


    For the grams of water produced
    1 mole of Hcl will produce 1 mole of water
    Let the number of gram of the water produced be g grams
    so 1= g/18
    g=18
    Therefore, 18 grams of water is produced

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