Answer to Question #151635 in General Chemistry for Sophie

Question

4.5 g of copper reacts with 1.50 L of 0.225 M nitric acid according to the following reaction. If the percent yield is 89.0%, what is the actual volume of nitrogen dioxide gas produced at 84.0 kPa and 42.0 degrees Celsius?

Answer 1

4 HNO₃(l) + Cu(s) ==> Cu(NO₃)₂(aq) + 2 NO₂(g) + 2 H₂O(l)

Moles of Cu=4.5/63.546=0.0708moles

Moles of HNO₃=1.5×0.225

                        =0.3375 moles

The moles that were all consumed in reaction are 0.0708moles of Cu

Moles of NO₂= (0.0708)4

=0.1416molesV₂=P₁V₁T₂/T₁P₂

At room temperature the 1mole=24dm³

                                              =0.1416 dm³

                                                 =24×0.1416

                                                   =3.3984 dm³

T₂=315K

1atm=101325Pa

%Yield of NO₂=0.89 ×3.3984

=3.02563 dm³

V₂=315×101325×3.0246/ (298×8400)

V₂=38.57 dm³