Answer to Question #114244 in General Chemistry for Daniel

Question

2.000 g of copper was reacted with 5.000 g of nitric acid (see reaction before Q1.) a. What is the limiting reagent? b. How many moles of copper(II) nitrate are formed? c. How many grams of the excess reagent are left unreacted?

Answer 1

The reaction can be shown as following:

4HNO₃ + Cu --> Cu(NO₃)₂ + 2NO₂ + 2H₂O

a) The limiting reactant can be calculated from the number of moles:

n(HNO₃) = m(HNO₃) / 4Mr(HNO₃) = 5.000 g / (4 × 63.01 g/mol) = 0.0198 mol

n(Cu) = m(Cu) / Mr(Cu) = 2.000 g / 63.546 g/mol = 0.0315 mol

As n(HNO₃) < n(Cu), nitric acid is a limiting reactant.

b) As nitric acid is a limiting reactant, 0.0198 mol of copper nitrate are formed.

c) Similarly, 0.0198 mol of copper reacted. As a result, 0.0315 mol - 0.0198 mol = 0.0117 mol of copper were left. From here:

m(Cu) = 0.0117 mol × 63.546 g/mol = 0.7435 g of copper were left unreacted.