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.
Step-by-step 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.