Question #86746

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Answer 1 · 2014-04-27T22:57:57

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

Step 1. Convert grams of ${CaC}_{2}$ $"CaC"_2$ to moles of ${CaC}_{2}$ $"CaC"_2$ .

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

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

$"CaC"_2 + "2H"_2"O" → "Ca(OH)"_2 + "C"_2"H"_2"$

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

Step 3. Use the Ideal Gas Law to calculate the volume of $"C"_2"H"_2$ at STP.

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

$PV = nRT$

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

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.