Explain how many moles of malic acid (c4h6o5)

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29 grams dry mix 28 grams of sugar a) What is the % by mass of the dry mix that is made up of "sugars"? (use 3 s.f. in all your answers for this problem) b) Malic acid is #2 on the ingredients list, so let\'s assume that 3/4 of the mix that is not \"sugars\" is malic acid. Using this estimate, what is the % by mass of malic acid in the mix? c) If you measured out 5.00 g of dry mix and used the estimated % malic acid calculated in step b, how many moles of malic acid (C4H6O5) would this contain? d) Using the reaction stoichiometry for the chemical reaction of malic acid with NaOH (aq) shown in your lab manual, calculate how many mL of 0.200 M NaOH (aq) would be needed to react with the malic acid sample from step c.

Reference no: EM13313774

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