Question

3. Consider the scenario where 50.00 mg of copper and 10 mL of 0.2 M nitric acid were mixed. The reac- tion that occurs is: Cu(s) + HNO3(aq) " Cu(NO3)2(aq) + NO2(g) + H2O(l). All of the gas formed from this reaction is collected over water in the same manner as the lab; the volume obtained was 20.52 mL. What percentage of the starting copper was consumed in this reaction? Assume room temperature (22.00°C) and atmospheric pressure (1.000 atm) for your calculations.

Answer 1

To determine the percentage of copper consumed in the given reaction, we need to calculate the moles of copper used and compare it to the initial moles of copper present.

First, let's calculate the moles of copper used in the reaction. The molar mass of copper (Cu) is 63.55 g/mol. We are given that 50.00 mg of copper is used, so we need to convert this mass to moles:

Mass of copper = 50.00 mg = 0.05000 g

Moles of copper = (mass of copper) / (molar mass of copper)

= 0.05000 g / 63.55 g/mol

≈ 0.000787 mol

According to the balanced equation, the stoichiometric ratio between copper and nitric acid is 1:2. This means that for every mole of copper consumed, two moles of nitric acid are required.

Therefore, the moles of nitric acid used in the reaction can be calculated as follows:

Moles of nitric acid = 2 * (moles of copper)

= 2 * 0.000787 mol

≈ 0.001574 mol

Now, let's determine the volume of NO2 gas collected over water. The volume obtained is given as 20.52 mL, but we need to correct it to standard temperature and pressure (STP) conditions for accurate calculations.

The STP conditions are defined as a temperature of 0°C (273.15 K) and a pressure of 1 atmosphere (atm). We are given that the room temperature is 22.00°C, so we need to convert it to Kelvin:

Temperature in Kelvin = Room temperature + 273.15

= 22.00°C + 273.15

≈ 295.15 K

Now, we can apply the ideal gas law to calculate the moles of NO2 gas produced:

PV = nRT

Where:

P = Pressure (in atm)

V = Volume (in liters)

n = Moles of gas

R = Ideal gas constant (0.0821 L·atm/mol·K)

T = Temperature (in Kelvin)

Rearranging the equation to solve for moles of gas:

n = (PV) / (RT)

Plugging in the values:

n(NO2) = (1.000 atm * 20.52 mL) / (0.0821 L·atm/mol·K * 295.15 K)

= 0.02052 L·atm / (0.0821 L·atm/mol·K * 295.15 K)

≈ 0.000876 mol

According to the balanced equation, the stoichiometric ratio between copper and NO2 is 1:1. This means that one mole of copper consumed corresponds to one mole of NO2 produced.

Therefore, the moles of copper consumed in the reaction are equal to the moles of NO2 produced:

Moles of copper consumed = Moles of NO2

≈ 0.000876 mol

To calculate the percentage of copper consumed, we need to compare the moles of copper consumed to the initial moles of copper present:

Percentage of copper consumed = (moles of copper consumed / initial moles of copper) * 100

= (0.000876 mol / 0.000787 mol) * 100

≈ 111.24%

Therefore, approximately 111.24% of the starting copper was consumed in this reaction.

Step-by-step explanation: