Grams of butane (C₄H₁₀) formed by the liquefaction of 448 litres of the gas (measured at (STP) would be

ANS:C - 1160

To find the mass of butane formed by the liquefaction of 448 liters of gas measured at STP (standard temperature and pressure), we need to consider the molar volume of the gas at STP. At STP, 1 mole of any ideal gas occupies approximately 22.4 liters. Given that the gas is butane (𝐶4𝐻10C4​H10​), we need to calculate the number of moles of butane formed from the given volume: Volume of gas=448 litersVolume of gas=448liters Molar volume at STP=22.4 liters/molMolar volume at STP=22.4liters/mol Number of moles of butane=Volume of gasMolar volume at STPNumber of moles of butane=Molar volume at STPVolume of gas​ Number of moles of butane=448 liters22.4 liters/molNumber of moles of butane=22.4liters/mol448liters​ Number of moles of butane=20 molesNumber of moles of butane=20moles Now, we need to find the molar mass of butane (𝐶4𝐻10C4​H10​): Molar mass of 𝐶4𝐻10=4×molar mass of 𝐶+10×molar mass of 𝐻Molar mass of C4​H10​=4×molar mass of C+10×molar mass of H Molar mass of 𝐶4𝐻10=4×12.01 g/mol+10×1.01 g/molMolar mass of C4​H10​=4×12.01g/mol+10×1.01g/mol Molar mass of 𝐶4𝐻10=48.04 g/mol+10.1 g/molMolar mass of C4​H10​=48.04g/mol+10.1g/mol Molar mass of 𝐶4𝐻10=58.14 g/molMolar mass of C4​H10​=58.14g/mol Now, to find the mass of butane formed: Mass of butane=Number of moles of butane×Molar mass of 𝐶4𝐻10Mass of butane=Number of moles of butane×Molar mass of C4​H10​ Mass of butane=20 moles×58.14 g/molMass of butane=20moles×58.14g/mol Mass of butane=1162.8 gMass of butane=1162.8g So, the mass of butane formed by the liquefaction of 448 liters of the gas measured at STP would be approximately 1162.8 grams. Therefore, the closest answer to 1162.8 grams from the given options is 1160 grams.