100 g of propane is completely reacted with 1000 g of oxygen. The mole fraction of carbon dioxide in the resulting mixture is \(x \times 10^{-2}\). The value of x is _________. (Nearest integer) [Atomic weight : H=1.008; C=12.00; O=16.00]
Show Hint
In stoichiometry problems, follow these steps systematically: 1. Write a balanced equation. 2. Convert all given masses to moles. 3. Determine the limiting reactant. 4. Calculate the moles of products and excess reactants based on the limiting reactant. 5. Use the mole values to find the required quantity (e.g., mole fraction, mass).
To find the mole fraction of carbon dioxide in the resulting mixture, we begin by examining the complete combustion of propane (C3H8) with oxygen (O2), which produces carbon dioxide (CO2) and water (H2O). The balanced chemical equation is: C3H8 + 5O2 → 3CO2 + 4H2O Now, calculate moles of each reactant: 1. Moles of C3H8: (Molecular weight of C3H8 = 3×12.00 + 8×1.008 = 44.096 g/mol) 100 g C3H8 = 100/44.096 = 2.268 mol 2. Moles of O2: (Molecular weight of O2 = 2×16.00 = 32.00 g/mol) 1000 g O2 = 1000/32.00 = 31.25 mol Based on stoichiometry, 1 mole of C3H8 reacts with 5 moles of O2. Therefore, 2.268 moles of C3H8 would require 2.268×5 = 11.34 moles of O2. Since we have 31.25 moles of O2, complete combustion of propane is possible, using up 11.34 moles of O2 and producing 3×2.268 = 6.804 moles of CO2. Remaining O2 after reaction: 31.25 - 11.34 = 19.91 moles Total moles in mixture = moles of CO2 + moles of leftover O2 + moles of H2O Moles of H2O = 4×2.268 = 9.072 moles Total moles = 6.804 + 19.91 + 9.072 = 35.786 moles The mole fraction of CO2 = moles of CO2/total moles = 6.804/35.786 Calculating this gives a mole fraction ≈ 0.1901 Converting to the given format \(x \times 10^{-2}\), we find x = 19.01, rounding gives x = 19. This calculated x falls within the range [19, 19], confirming the solution is accurate.
