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QUESTION

# Calcium carbide reacts with water to produce acetylene gas. CaC_(2(s))+2H_2O_((l)) -&gt; C_2H_(2(g)) + Ca(OH)_(2(g) What volume of acetylene at 25°C and 1.01 atm is generated by the complete reaction of 2.49g of calcium carbide? What is the volume at STP...

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 to moles of "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.