Final answer:
Approximately 0.9282 grams of copper are needed to completely react with 2.45 grams of nitric acid, using stoichiometry to find the molar ratio from the balanced chemical equation.
Step-by-step explanation:
The student is asking about the amount of copper that would completely react with a given amount of nitric acid (HNO3). In order to solve this problem, it's essential to use stoichiometry from a balanced chemical equation which involves copper (Cu) and nitric acid reacting to produce copper(II) nitrate (Cu(NO3)2) and other products depending on the concentration of the nitric acid used. The balanced chemical equation for this reaction, if using concentrated nitric acid, is:
3Cu(s) + 8HNO3(aq) → 3Cu(NO3)2(aq) + 2NO(g) + 4H2O(l)
From this equation, we can determine the molar ratio between copper and nitric acid to be 3 mol of Cu: 8 mol of HNO3. Since we are starting with a mass of nitric acid, the first step is to convert the mass of nitric acid to moles using its molar mass.
The molar mass of HNO3 is approximately 63.01 g/mol. Therefore, 2.45 grams of HNO3 is:
2.45 g HNO3 × (1 mol HNO3 / 63.01 g) ≈ 0.0389 mol HNO3
Using the molar ratio from the balanced equation, we calculate the moles of copper:
0.0389 mol HNO3 × (3 mol Cu / 8 mol HNO3) ≈ 0.0146 mol Cu
Finally, we convert the moles of copper back to grams using the molar mass of copper which is approximately 63.55 g/mol:
0.0146 mol Cu × (63.55 g / 1 mol Cu) ≈ 0.9282 grams of Cu
This calculation results in approximately 0.9282 grams of copper needed to react completely with 2.45 grams of nitric acid.