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150 cm3 0.24 M naoh solution react with excess al(no3)3, calculate the mass of saodium aluminate produced.

al(no3)3 + 4naoh -> naalo2 + 3 nano3 +2h20

i got 0.738g..
is my answer correct? if im wrong could u show me the calculation?

  • chem -

    Please use conventional symbols as this makes things easier to read and you also practise with the 'language' of the subject.

    Al(NO3)3 + 4NaOH -> NaAlO2 + 3 NaNO3 +2H2O

    So 4 moles of NaOH produce 1 moles of NaAlO2.
    or
    So 1 mole of NaOH produce 1/4 moles of NaAlO2.


    We start with

    0.150 L x 0.24 mole L^-1 = 0.036 mole

    this must produce

    0.036 mole/4 moles of NaAlO2
    =0.009 moles

    The relative molecular mass of NaAlO2 is

    23+27+32 = 82

    so the mass produced is

    0.009 moles x 82 g mole^-1

    =0.738 g

    so you are correct.

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