Determine the mass of sodium nitrate produced when 0.73 g of aqueous nickel(II) nitrate reacts with aqueous sodium hydroxide. How would you recover this nitrate as a solid from the reaction?
Determine the mass of sodium nitrate produced when 0.73 g of aqueous nickel(II) nitrate reacts with aqueous sodium hydroxide. How would you recover this nitrate as a solid from the reaction?
NOTE : There is a typing error in question. The solid will be of nickel (II) hydroxide and and sodium nitrate.
And you will recover nickel (II) hydroxide from the reaction produced by the nickel (II) nitrate.
Given : Mass of nickel (II) nitrate reacting = 0.73 g
Molar mass of nickel (II) nitrate i.e Ni(NO3)2 = Atomic mass of Ni + Atomic mass of N X 2 + Atomic mass of O X 6 = 58.7 + 14 X 2 + 16 X 6 = 182.7 g/mol.
Since mass = moles X molar mass
=> 0.73 = moles of Ni(NO3)2 reacting X 182.7
=> Moles of Ni(NO3)2 reacting = 0.0040 mol approx.
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