Calcium carbide reacts with water to produce acetylene gas, C2H2. Calculate the volume (in liters) of acetylene produced at 25.0 °C and 685 mmHg from 0.0750 mol CaC2 and excess H2O. Given that R = 0.0821 L-atm/(mol-K). CaC2(s) + 2H2O(l) -> Ca(OH)2(aq) + C2H2(g) a. 4.75 L. b. 2.04 L. c. 3.39 L. d. 2.72 L. e. 4.07 L.
The Correct Answer and Explanation is:
To determine the volume of acetylene gas (C₂H₂) produced, we use the Ideal Gas Law: PV=nRTPV = nRT
Where:
- PP = pressure in atm
- VV = volume in liters
- nn = moles of gas
- RR = ideal gas constant = 0.0821 L·atm/(mol·K)
- TT = temperature in Kelvin
Step 1: Identify the amount of gas produced
From the balanced equation: CaC2(s)+2H2O(l)→Ca(OH)2(aq)+C2H2(g)\text{CaC}_2(s) + 2\text{H}_2\text{O}(l) \rightarrow \text{Ca(OH)}_2(aq) + \text{C}_2\text{H}_2(g)
1 mole of CaC₂ produces 1 mole of C₂H₂ gas.
Given: 0.0750 mol CaC₂ → produces 0.0750 mol of C₂H₂.
Step 2: Convert pressure to atm
685 mmHg×1 atm760 mmHg=0.901atm685 \text{ mmHg} \times \frac{1 \text{ atm}}{760 \text{ mmHg}} = 0.901 atm
Step 3: Convert temperature to Kelvin
T=25.0°C+273.15=298.15KT = 25.0°C + 273.15 = 298.15 K
Step 4: Plug values into the Ideal Gas Law
V=nRTP=(0.0750 mol)(0.0821 L\cdotpatm/mol\cdotpK)(298.15 K)0.901 atmV = \frac{nRT}{P} = \frac{(0.0750 \text{ mol}) (0.0821 \, \text{L·atm/mol·K}) (298.15 \text{ K})}{0.901 \text{ atm}} V=1.834 L\cdotpatm0.901 atm≈2.04 LV = \frac{1.834 \, \text{L·atm}}{0.901 \, \text{atm}} \approx 2.04 \, \text{L}
✅ Correct Answer: b. 2.04 L
Explanation
In this problem, we calculate the volume of acetylene gas (C₂H₂) produced from a chemical reaction between calcium carbide (CaC₂) and water. The balanced chemical reaction shows that one mole of calcium carbide produces one mole of acetylene gas, assuming excess water is present.
We start with 0.0750 mol of CaC₂. This directly gives 0.0750 mol of C₂H₂ since the mole ratio is 1:1. Next, we apply the Ideal Gas Law equation PV=nRTPV = nRT, which relates pressure (P), volume (V), temperature (T), and the number of moles (n) for an ideal gas.
To use this formula, the pressure must be in atmospheres (atm), and the temperature must be in Kelvin (K). The pressure is given in mmHg, so we convert 685 mmHg to atm using the conversion factor 1 atm=760 mmHg1 \text{ atm} = 760 \text{ mmHg}, which results in 0.901 atm. Similarly, the temperature is converted from Celsius to Kelvin by adding 273.15, resulting in 298.15 K.
Substituting these values into the Ideal Gas Law formula with the gas constant R=0.0821 L\cdotpatm/(mol\cdotpK)R = 0.0821 \text{ L·atm/(mol·K)}, we solve for volume VV. After calculations, the result is approximately 2.04 liters.
This calculation assumes ideal gas behavior, which is a good approximation under many standard conditions like 25°C and around 1 atm pressure. The final volume reflects how much space 0.0750 mol of acetylene gas occupies under those conditions.
Thus, the correct answer is b. 2.04 L.
