If 1.000 g copper wire were reacted wtih 10.00 mL of concentrated nitric acid (1
ID: 1002015 • Letter: I
Question
If 1.000 g copper wire were reacted wtih 10.00 mL of concentrated nitric acid (16 M), which is the limiting reactant? How many moles of copper (II) nitrate are produced? What percentage of the non-limiting reactant remains? Please put 1 answer per blank. blank1: chemical formula for the limiting reactant blank 2: numeric answer for moles of product produced (remember to use to correct significant figures!) blank 3: unit (abbreviation/symbol only) for the numeric answer in blank 2 blank 4: numeric answer for percent of non-limiting reactant remaining (remember to use correct significant figures) blank 5: unit (abbreviation/symbol only) for the numeric answer in blank 4Explanation / Answer
Cu + 4 HNO3 Cu(NO3)2 + 2 H2O + 2 NO2
(Note that this balanced equation is for concentrated HNO3. Dilute HNO3 would produce NO gas instead.)
(1.000 g Cu) / (63.5463 g Cu/mol) = 0.01574 mol Cu
0.01574 moles of cooper nitrate is produced as Cu is limiting reagent
(0.01000 L) x (16 mol/L) = 0.16 mol HNO3
0.01574 mole of Cu would react completely with 4 x 0.01574 = 0.06296 mole of HNO3, but there is more HNO3 present than that, so HNO3 is in excess and Cu is the limiting reactant.
0.16 mol HNO3 - 0.06296 mol HNO3used = 0.09704 mol HNO3 remaining
0.09704 / 0.16 = 0.6065 = 60.65% of the original HNO3 remains Cu + 4 HNO3 Cu(NO3)2 + 2 H2O + 2 NO2
blank1: chemical formula for the limiting reactant blank (Cu)
blank 2: numeric answer for moles of product produced (remember to use to correct significant figures!) = 0.01574 moles
blank 3: unit (abbreviation/symbol only) for the numeric answer in blank 2 = moles
blank 4: numeric answer for percent of non-limiting reactant remaining (remember to use correct significant figures) = 60.65%
blank 5: unit for the numeric answer in blank 4 = % (no units)
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