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    Stoichiometry

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    6) Explain why the reaction of 31.1636 g of aluminum metal (atomic mass 26.9815 g/mol) with 110.00 g of sodium iodate (molar mass 197.89 g/mol) in the presence of nitric acid produces the equivalence of only 25.00 g of aluminum ions; whereas,reaction with 120.00 g of sodium iodate produces the equivalence of 31.1636 g of aluminum ions.

    7) You place 129.50 g of sodium hydroxide into exactly 1635 mL of water. With how many mL of 1.9801 M hydrochloric acid need to be added to react completely (i.e.,titrate) the sodium hydroxide? What will the final concentration of the sodium chloride be in the mixture?

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    https://brainmass.com/chemistry/reaction-stoichiometry/stoichiometry-reactions-aluminum-sodium-hydroxide-249439

    Solution Preview

    Please help me with the following stoichiometric problems
    6) Explain why the reaction of 31.1636 g of aluminum metal (atomic mass 26.9815 g/mol) with 110.00 g of sodium iodate (molar mass 197.89 g/mol) in the presence of nitric acid produces the equivalence of only 25.00 g of aluminum ions; whereas,reaction with 120.00 g of sodium iodate produces the equivalence of 31.1636 g of aluminum ions.
    The key here is looking at how much of the aluminum is converted. In the first reaction, 31.1636 g of Al are converted to 25.00 g of ...

    Solution Summary

    Why an excess of sodium iodate will produce more aluminum ions than a smaller amount of sodium iodate is determined. The expert determines how to calculate the volume of HCl needed to neutralize a specific mass of NaOH.

    $2.19