Q2. A 0.696 mol sample of Cu is added to 146 mL of 5.5 M HNO3. Assuming the following reaction is only one that occurs;
Cus) + HNO3(aq) → Cu(NO3)2(aq) + H2O() + NO()
Will the Cu react completely? What is the limiting reagent and what is the remaining compound
in mass?

The limiting reagent is HNO3 and the remaining compound in mass is 25.1 grams of Cu.

The reaction taking place in the given case is,

Based on the given information,

•The moles of Cu given is 0.696 moles.  

•The volume of HNO3 given is 146 ml and the molarity of HNO3 is 5.5 M.

Now the moles of HNO3 can be determined as,

•In order to find the limiting reagent, there is a need to divide moles by their stoichiometric coefficient, that is,  

Cu = 0.696/3 = 0.232 moles and HNO3 = 0.803/8 = 0.100 moles

•As the moles of HNO3 is present in less amount, therefore, HNO3 will be the limiting reagent.  

Now the remaining copper left in the reaction will be,

Now the mass of remaining copper will be = Moles * Molecular mass of Cu

Thus, the mass of remaining compound is 25.1 grams and the limiting reagent is HNO3.

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Explanation:

Given parameters:

Number of moles of Cu  = 0.696mol

Volume of HNO₃  = 146mL

Concentration of HNO₃  = 5.5M

Unknown:

Will Cu react completely = ?

Limiting reagent  = ?

Remaining mass of the compound  = ?

Solution:

The equation of the reaction is given as;

             Cu + 4HNO₃  →  Cu(NO₃)₂  + 2H₂O  + 2NO₂

Let us find the number of moles of HNO₃ ;

  Number of moles  = concentration x volume

  Number of moles  = 5.5 x 146 x 10⁻³L = 0.803mol

From the balanced reaction equation;

      1 mole of Cu reacted with 4 mole of HNO₃

       0.696 mole of Cu will require 4 x 0.696  = 2.784mole of HNO₃

So, the limiting reactant is HNO₃ because it is in short supply.

Cu will completely react

Mass of the remaining compound we need;

       mass  = number of moles x molar mass

  Number of moles = 2.784  - 0.803  = 1.981mole

So, mass  = 1.981 x (1 + 14 + 3(16))  = 124.81g of HNO₃ is required to make the reaction complete.

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percent abundance for cu-63 can be shown as:

63 x + 65 (1-x) = 62.9296

65 - 62.9296 = 2 x

x = 1.035

thus, percent abundance for cu-63 is 1.035

percent abundance for cu-65 can be shown as:

63 x + 65 (1-x) = 64.9278

65 - 64.9278 = 2 x

x = 0.0361

thus, percent abundance for cu-63 is 0.0361

the mass spectrum of cu-63 showing the intense peak as, this isotope is found in nature more abundantly around 92.38 % in comparison to cu-65 which is found 62.58 % naturally