Question

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?

a) 1 drop
b) 2 drops
c) 3 drops
d) 4 drops

Answer 1

To react with 0.030 g of copper metal, 0.067 mL of 15 M nitric acid is required, which is approximately 1 drop when considering the ratio of drops to milliliters.

Step-by-step explanation:

To determine the volume of 15 M nitric acid required to react with 0.030 g of copper metal, we first need to balance the chemical equation between copper and nitric acid. The balanced equation is as follows:

3Cu + 8HNO3 → 3Cu(NO3)2 + 2NO + 4H2O

From the balanced equation, we can see that the ratio of copper to nitric acid is 3:8. Therefore, if 0.030 g of copper requires 8 parts of nitric acid, we can calculate the volume of nitric acid needed as follows:

0.030 g Cu × (8/3) × (1 mol Cu/63.546 g Cu) × (1 L/15 mol HNO3) × (1000 mL/1 L) = 0.067 mL of 15 M nitric acid

Since 1.0 mL of acid contains approximately 20 drops, we can calculate the number of drops needed:

0.067 mL × (20 drops/1 mL) ≈ 1 drop

Therefore, the answer is a) 1 drop.