Question

Suppose that you have a solid sample only containing the element Quakerium (Qu) and a small amount of Locust Carbonate (LoCO3). You add HCl to the sample, and both substances react completely with excess hydrochloric acid. The resulting solution (25.0 mL in total) after the chemical reaction is "Penn Blue" in color with colorless gases evolving (although you suspect there may be 2 different gases). There are no precipitates and the solution is clear. There are two chemical equations representing the reactions. Chemical notes/hints: Qu+ (aq) is a Penn Red and Qu2+ (aq) is colorless. All ions of Locust are Penn Blue in solution.

1. What are all the products in the reaction between Qu (s) and HCl (aq)? Select all.
HCl (aq)
QuCO3 (aq)
LoCO3 (aq)
H2 (g)
QuCl (aq)
QuCl2 (aq)
LoCl2 (aq)
LoCl (aq)
H2O (l)
CO2 (g)
CO3 (g)
HCO3 (g)
H2CO3 (g)

2. What are the products in the reaction between LoCO3 (s) and HCl (aq)? Select all of the products.
HCl (aq)
QuCO3 (aq)
LoCO3 (aq)
H2 (g)
QuCl (aq)
QuCl2 (aq)
LoCl (aq)
LoCl2 (aq)
H2O (l)
CO2 (g)
CO3 (g)
HCO3 (g)
H2CO3 (g)

3. You decide to use absorbance to measure the resulting solution. You dilute 2.00 mL of the 25.0 mL solution with water in a 100 mL volumetric flask. The absorbance of the new diluted solution is 0.85. If the equation of the trendline from the standardization plot was y=1,106x, determine the mass of the LoCO3 in the original sample. The MW of LoCO3 is 128.52 g/mol.

4. Using the moles of Locust Carbonate calculated in the question above, calculate the volume of gas produced in the reaction with Locust Carbonate. Assume the gas behaves ideally and you are at standard temperature (T=293 K) and pressure (P=1.0 atm).


5. Given that the total volume of gas produced in the complete reaction (both reactions combined) is 0.500 L, calculate the mass of Quakerium in the sample (Assume Qu is 34.02 g/mol and you are at standard temperature (T=293 K) and pressure (P=1.0 atm)).

Answer 1

These answers provide the products for each reaction, calculate the mass of Locust Carbonate from absorbance, and determine the volume and mass of gases based on ideal gas behavior. Remember, these calculations rely on several assumptions, and the actual values may slightly deviate.

1. Products of Qu + HCl:

H2 (g): Quakerium reacts with HCl to produce hydrogen gas.

QuCl2 (aq): Quakerium forms the Qu2+ ion, which reacts with Cl- to form QuCl2.

H2O (l): HCl provides the H+ for the reaction with Qu and Cl- for the QuCl2 formation, creating water molecules.

2. Products of LoCO3 + HCl:

CO2 (g): Locust Carbonate releases CO2 gas when reacted with HCl.

LoCl2 (aq): The carbonate ion forms Lo2+ and combines with Cl- to create LoCl2 in solution.

H2O (l): Water is a byproduct of the acid-base reaction with the carbonate ion.

3. LoCO3 Mass from Absorbance:

Absorbance 0.85 for 2 mL diluted solution indicates the original concentration of Lo2+ was 0.85/2 = 0.425 M in the 25 mL solution.

Using the trendline equation, the concentration corresponds to
`(0.425 )/( 1,106) = 3.84 x 10^(-4 )`mol LoCO3 in 25 mL.

Multiplying by the original volume and molar mass, the mass of LoCO3 is
`3.84 x 10^(-4) mol * 25 mL * 128.52` g/mol = 0.123 g.

4. Volume of LoCO3-derived Gas:

Assuming ideal gas behavior and standard conditions, the volume of CO2 from
`3.84 x 10^(-4)` mol is calculated using the ideal gas law:
`V = nRT/P = 3.84 x 10^(-4)` mol * 0.0821 L atm/mol K * 293 K / 1 atm ≈ 0.09 mL.

5. Quakerium Mass from Total Gas Volume:

Subtracting the LoCO3-derived gas volume from the total volume gives 0.500 L - 0.09 mL ≈ 0.499 L of gas from Quakerium.

Using the ideal gas law and molar mass of Quakerium, the mass of Quakerium is calculated as: M = nRT/P = 0.499 L / (0.0821 L atm/mol K * 293 K * 1 atm) * 34.02 g/mol ≈ 0.185 g.