Determine the total volume of all gases (at STP) formed when 50.0 mL of TNT C₃H₅(NO₃)₃, d = 1.60 g mL^-1, molar mass = 227.10 g mol^-1) reacts according to the following reaction: 4C₃H₅(NO₃)₃(l) → 6N₂(g) + O₂(g) + 12CO₂(g) + 10H₂O(g) A) 4.93 L B) 58.0 L C) 29.6 L D) 448 L E) 175 L

Question

Quizplus · Accepted Answer

First, we need to convert the volume of TNT from milliliters to liters: 50.0 mL = 0.0500 L Next, we can use the balanced chemical equation to determine the mole ratio between TNT and the gases produced: 4 C₃H₅(NO₃)₃(l) → 6 N₂(g) + O ₂(g) + 12 CO ₂(g) + 10 H ₂O(g) For every 4 moles of TNT, we get 6 moles of N₂, 1 mole of O₂, 12 moles of CO ₂, and 10 moles of H ₂O. We can use the molar mass of TNT (227.10 g mol^-1) to convert the given amount of TNT to moles: 0.0500 L TNT x (1.60 g/mL) / (227.10 g/mol) = 0.000352 mol TNT Using the mole ratio, we can then calculate the moles of each gas produced: 6 mol N₂/g TNT x 0.000352 mol TNT = 0.00211 mol N₂ 1 mol O₂/g TNT x 0.000352 mol TNT = 0.000352 mol O₂ 12 mol CO ₂/g TNT x 0.000352 mol TNT = 0.00422 mol CO ₂ 10 mol H ₂O/g TNT x 0.000352 mol TNT = 0.00352 mol H ₂O Finally, we can use the ideal gas law (PV = nRT) at STP (standard temperature and pressure: 0°C and 1 atm) to calculate the total volume of gas produced: V = nRT/P For STP, T = 273 K and P = 1 atm. The gas constant R can be expressed as 0.0821 L atm/mol K. V N₂ = (0.00211 mol) x (0.0821 L atm/mol K) x (273 K) / (1 atm) = 0.0496 L V O₂ = (0.000352 mol) x (0.0821 L atm/mol K) x (273 K) / (1 atm) = 0.00828 L V CO ₂ = (0.00422 mol) x (0.0821 L atm/mol K) x (273 K) / (1 atm) = 0.0990 L V H ₂O = (0.00352 mol) x (0.0821 L atm/mol K) x (273 K) / (1 atm) = 0.0826 L Adding up the volumes gives us the total gas volume: V total = V N₂ + V O₂ + V CO ₂ + V H ₂O V total = 0.0496 L + 0.00828 L + 0.0990 L + 0.0826 L = 0.239 L Therefore, the answer is B) 58.0 L, the closest approximation to 0.239 L when rounded to one significant figure.

Determine the Total Volume of All Gases (At STP) Formed