Determining the Molecular Formula

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Limiting ReactantsReaction will stop when one of the reactantsis gone3H (g) + N (g) → 2NH (g)2 2 3How many moles of product will result from3.0 mol N 2 and 6.0 mole H 23.0 mol N 2 can use 9.0 mol H 2 , have 6.06.0 mol H 2 can use 2 mol N 2 , have 3.0All H 2 will be consumed and so it is thelimiting reactant
Given the reaction2Na PO ( aq ) + 3Ba( NO ) ( aq ) → Ba ( PO ) ( s ) + 6NaNO ( aq )3 4 3 2 3 4 2 3350 . g Na3PO 4and 6.40g Ba(NO3) 2How many grams of Ba3(PO 4)2will be formed?Which species is the limiting reactant?Compound3 43 2MolesNa PO 0.0213Ba(NO ) 0.0245Ba(NO3) 2is limiting reactantMoles of Ba (PO ) formed3 4 21 moles of Ba(NO3 )2 reacted3which is 0.00816
Page 1 and 2: Determining the Molecular Formula1.
Page 3 and 4: Molecular formula from combustionGi
Page 5: moles O =0.061g16.00g/mol =0.0038 m
Page 9 and 10: Dilution ProblemGiven a solution wi

Determining the Molecular Formula

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