- 0.6 mole of copler 2 sulphade LS Treated with 1.40 mole of trioxonitrate (v) acid According to - Question 1) How many mole of copper 2 ultrate could be formed 2) What is the Limiting Reagent In this Reaction 3) Calculate the Clumber of nole of excess reagent left at the end of the readion Solution

To solve this reaction, we first balance the equation between copper(II) sulfate (CuSO₄) and nitric acid (HNO₃). The balanced equation typically shows that copper(II) sulfate reacts with nitric acid to form copper(II) nitrate (Cu(NO₃)₂), water (H₂O), and sulfuric acid (H₂SO₄) in a known mole ratio. Based on the balanced equation, you'd find the limiting reagent by comparing the mole ratios. If you start with 0.6 moles of CuSO₄ and 1.40 moles of HNO₃, calculate how much of each reactant is needed for the complete reaction, and determine which one you have less of based on stoichiometry. This will give you the limiting reagent. If CuSO₄ reacts according to a ratio of 1:2 with nitric acid, you would need 1.2 moles of HNO₃ (0.6 moles CuSO₄ x 2), leaving an excess of HNO₃. The excess reagent left can then be calculated by subtracting the moles used from the original moles you had of HNO₃. Subtract 1.2 moles (used) from 1.40 moles (initial), which leaves you with 0.20 moles of the excess reagent (HNO₃).