If 1.000 g copper wire were reacted with 10.00 mL of concentrated nitric acid (1

Question

If 1.000 g copper wire were reacted with 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. blank 1: 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 4

Explanation / Answer

net reaction

3Cu + 8HNO3 --------> 3Cu(NO3)2 + 2NO + 4H2O

here no of mole of Cu used = given mass/ atomic mass of Cu = 1/63 =0.0158 moles

with HNO3 used is 16M we can calulate no of moles present using

Volume of solution used 10.0 mL = 0.010L

Molarity = no of moles/ volume of solution in litres

thus no of moles = Molarity * volume of solution = 16 * 0.01 = 0.16 moles

thus from above equation we can conclude

3 moles of copper reacts with 8 moles of HNO3 to form forms 3 moles of Copper nitrate

thus 0.0158 moles of copper will react with x moles of nitratic acid = 0.0158 * 8 /3 = 0.0421 moles of nitric acid

we can say 0.0158 moles of copper require 0.0421 moles of nitric acid for complete reaction, but the nitric acid used was 0.16 moles. thus nitric acid will be left in the reaction unreacted after the completion of reaction

(a) thus our limiting reactant is Copper - Cu

(b) moles of product produced will be 0.0158 moles

as 3 moles of Cu reacts with HNO3 forming 3 moles of copper nitrtate,

thus 0.0158 moles of Cu will form = 0.0158 * 3 /3 = 0.0158 moles of copper nitrate

(c) unit is mol

(d) amount of non limiting Reactant , HNO3 = total no of moles taken - no of moles consumed during reaction

amount of non limiting Reactant , HNO3 = 0.16 - 0.0421 = 0.1179 moles

% of Non limiting reactant remaining = amount left unreacted *100 / total amount used = 0.1179 *100 / 0.16 = 73. 67 % is of HNO3 remained unreacted

(e) unit is %

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.