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(0.75 moles Al(NO3)3/L Al(NO3)3) x 0.040 L Al(NO3)3 = ??moles Al(NO3)3

I am unsure how to find the factor to convert moles Al(NO3)3 into moles NO3.

i think that if i do the first section,
you get .03 moles as your answer for moles of Al(NO3)3, but I am not sure.
how can i convert the correct answer for moles Al(NO3)3 into moles NO3 ?

  • Chemistry -

    The question said to use dimensional analysis...

  • Chemistry -

    You want NO3^- so how many NO3^- are in 1 mole Al(NO3)3. Three? Then moles NO3^- =
    ??moles Al(NO3)3 x [3 moles NO3^-/1 mol Al(NO3)3]
    Look how the units you don't want to keep cancel and the unit you want to keep stays.

  • Chemistry -

    oh ok thanks i got .09 i hope that's right

  • Chemistry -

    Yes and no. Some profs would count that as incorrect. Note there are two significant figures in 0.75 and 0.40; therefore, you are allowed two s.f. in the answer. The answer should be quoted as 0.090 and not 0.09 moles NO3^-.

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