Please answer question 18 part A Liquefied petroleum (LP) gas burns according to thefollowing exothermic reaction:You may want to reference (Pages 385 - 387) Section 9.7 while completing this problem.C3 Hs (g) + 502 (g) → 3CO2(g) + 4H2O(g)A Hom = -2044 kJPart AWhat mass of LP gas is necessary to heat 1.7 L of water from room temperature (25.0 ° C) to boiling (100.0°C)?Assume that, during heating, 16% of the heat emitted by the LP gas combustion goes to heat the water. The rest islost as heat to the surroundings.Express your answer using two significant figures.n ΑΣφ?m =

Given: The specific heat capacity of water, (C)= 4.18 J/g0C The volume of water, (V) = 1.7 L= 1700 g. The initial temperature of the water, (T1) = 25 0C. The final temperature of the water, (T2) = 100 0C. The density of water, (d) = 1 g/mL. The combustion reaction of the LP is. C3H8(g)+5O2(g)---------------->3CO2(g)+4H2O(g) ∆HRxn0=-2044 kJ According to the stoichiometry of the reaction, 1 mol of C3H8 burns to produce = 2044 kJ of heat.The mass of water, (m) used will be calculated as follows: mH2O=V×dmH2O=1700 g×1 g/mLmH2O=1700 gThe heat is required to raise the temperature of water from T1 to T2 is calculated as follows: Q=mH2O×C×T2-T1.............1Plugin the values in equation 1Q=1700 g×4.18 J/g0C×100-250CQ=532950 J=532.95 kJ Now, 16 % of total heat,(H) = 532.95 kJH= 532.95×10016H=3330.94 kJ.According to the stoichiometry of the reaction, 2044 kJ of heat is produced from the combustion of = 1 mol of C3H8 1 kJ of heat is produced from the combustion of = 12044 mol of C3H8 3330.94 kJ of heat is produced from the combustion of = 12044×3330.94 mol of C3H8 = 1.629 mol of C3H8The mass of C3H8 used to heat the water will be, mC3H8=moles×Molar massmC3H8=1.629×44.10 g/molmC3H8=71.8≈72 gThe massin grams of C3H8 used to heat the water = 72 g (Answer)