# Question 86746

Apr 27, 2014

The volume at STP is 882 mL. The volume at 25 °C and 1.01 atm is 941 mL.

#### Explanation:

Step 1. Convert grams of ${\text{CaC}}_{2}$ to moles of ${\text{CaC}}_{2}$.

2.49 cancel("g CaC₂") × ("1 mol CaC"_2)/(64.10 cancel("g CaC₂")) = "0.038 85 mol CaC"_2

Step 2. Use the molar ratio from the balanced equation to convert moles of $\text{CaC"_2}$ to moles of ${\text{C"_2"H}}_{2}$.

$\text{CaC"_2 + "2H"_2"O" → "Ca(OH)"_2 + "C"_2"H"_2}$

0.038 85 cancel("mol CaC₂") × ("1 mol C"_2"H"_2)/(1 cancel("mol CaC₂")) = "0.038 85 mol C"_2"H"_2

Step 3. Use the Ideal Gas Law to calculate the volume of ${\text{C"_2"H}}_{2}$ at STP.

STP is defined as a pressure of 100 kPa and a temperature of 273.15 K.

$P V = n R T$

V = (nRT)/P = (0.038 85 cancel("mol") × 8.314 cancel("kPa")"·L·"cancel("K⁻¹mol⁻¹") × 273.15cancel("K"))/(100 cancel("kPa")) = "0.882 L" = "882 mL"

The volume at STP is 882 mL.

Step 4. Use the Ideal Gas Law to calculate the volume at 25 °C and 1.01 atm.

V = (nRT)/P = (0.038 85 cancel("mol") × 0.082 06 "L·"cancel("atm·K⁻¹mol⁻¹") × 298.15cancel("K"))/(1.01 cancel("atm")) = "0.941 L" = "941 mL"#

The volume at 25 °C and 1.01 atm is 941 mL.