Questions: Question 1 (5 points) Review question #1-1: Consider the following redox reaction: Cu(s) + 4 HNO3(aq) à Cu(NO3)2(aq) + 2 NO2(g) + 2 H2O(l) If 5.48 g of Cu are allowed to react with excess HNO3, what is the theoretical yield (g) of Cu(NO3)2 in grams?

Solution Steps

• The molar mass of copper (Cu) is approximately \( 63.55 \, \text{g/mol} \).

• Use the formula: \[ \text{moles of Cu} = \frac{\text{mass of Cu}}{\text{molar mass of Cu}} \]
• Given mass of Cu = 5.48 g, calculate: \[ \text{moles of Cu} = \frac{5.48 \, \text{g}}{63.55 \, \text{g/mol}} \]

• From the balanced equation, 1 mole of Cu produces 1 mole of \(\mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}\).
• Therefore, moles of \(\mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}\) = moles of Cu.

• Calculate the molar mass:
  • Cu: \( 63.55 \, \text{g/mol} \)
  • N: \( 14.01 \, \text{g/mol} \times 2 = 28.02 \, \text{g/mol} \)
  • O: \( 16.00 \, \text{g/mol} \times 6 = 96.00 \, \text{g/mol} \)
• Total molar mass of \(\mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}\) = \( 63.55 + 28.02 + 96.00 = 187.57 \, \text{g/mol} \)

• Use the formula: \[ \text{mass of } \mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2} = \text{moles of } \mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2} \times \text{molar mass of } \mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2} \]
• Substitute the values to find the theoretical yield.

Final Answer