Question

6. In Reaction A, students are instructed to add no more than 0.25 mL of 15 M nitric acid. What volume of 15 M nitric acid is required to react with 0.030 g of copper metal

Answer 1

The given question is incomplete, the complete question is:

In Reaction A, students are instructed to add no more than 0.25 mL of 15 M nitric acid. What volume of 15 M nitric acid is required to react with 0.030 g of copper metal? If 1.0 mL of acid contains approximately 20 drops, how many drops of nitric acid are needed?

Answer:

The correct answer is 0.0629 ml and 1.26 drops.

Explanation:

Based on the given question, the equation is:  

Cu + 2HNO₃ (aq) ⇒ Cu(NO₃)₂ + H₂

The mass of copper given is 0.030 grams.  

The molecular mass of copper is 63.55 gram per mole. The number of moles can be determined by using the formula, n = weight/molecular mass.  

Moles of Cu = 0.030 grams/63.55 grams per mole = 0.000472 moles

Based on the reaction, it is clear that 1 mole of Cu reacts with 2 moles of nitric acid, therefore, the number of moles of nitric acid needed will be,  

= 0.000472 mol Cu × 2 mol HNO₃ / 1 mole Cu

= 0.000944 mol HNO₃

The concentration of HNO₃ given is 15 M

Now the volume of HNO₃ required to react with 0.030 grams of copper metal will be,  

Volume = 0.000944 mol HNO₃ × 1L/15 mol HNO₃ × 1000 ml/ 1L

= 0.0629 ml.  

Based on the given information, if 1 ml of nitric acid comprise 20 drops, therefore, 0.0629 ml of the acid will require the drops,  

Number of drops of HNO₃ = 0.0629 ml × 20 drops / 1 ml  

= 1.26 drops.