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) 0.0125 mL, 1 drop

B) 0.0375 mL, 2 drops

C) 0.0500 mL, 3 drops

D) 0.0750 mL, 4 drops

E) 0.1000 mL, 5 drops

Answer 1

To react with 0.030 g of copper, the required volume of 15 M nitric acid is approximately 0.0100 mL. Since 1.0 mL of acid contains about 20 drops, we need roughly 0.2 drops. Therefore, the correct answer is about 1 drop of acid, which corresponds to 0.0125 mL.

Step-by-step explanation:

To determine the volume of nitric acid required to react with copper metal, we must first use the stoichiometry of the reaction between copper and nitric acid, which is:

3 Cu(s) + 8 HNO3(aq) → 3 Cu(NO3)2(aq) + 2 NO(g) + 4 H2O(l)

From the chemical equation, we see that 3 moles of copper react with 8 moles of HNO3. Therefore, to react with 0.030 g of copper, we first find the moles of copper using its molar mass (63.55 g/mol).

Moles of Cu = mass / molar mass = 0.030 g / 63.55 g/mol = 4.72 x 10−5 mol

Next, we find the moles of HNO3 needed using the stoichiometry:

Moles of HNO3 = (8/3) x moles of Cu = (8/3) x 4.72 x 10−5 mol

From there, we calculate the volume of 15 M nitric acid required:

Volume of HNO3 = moles of HNO3 / concentration = (8/3 x 4.72 x 10−5) mol / 15 M

Volume of HNO3 ≈ 0.0100 mL

Given 1.0 mL of acid contains approximately 20 drops:

Drops of HNO3 = volume / drop volume = 0.0100 mL / (1.0 mL / 20 drops)

Drops of HNO3 ≈ 0.2 drops

Since we cannot have a fraction of a drop, we'll round up to the nearest whole drop, which is 1 drop. Therefore, the correct answer is 0.0125 mL, 1 drop, which corresponds to option A.