ANS:B - 10

To solve this problem, we need to calculate the amount of acetylene gas produced by 1 kg of calcium carbide and then determine how many hours of service can be derived from this amount of gas. Given:

• 1 kg of calcium carbide produces about 0.41 kg of acetylene gas.

First, we need to find out how much acetylene gas can be produced from 1 kg of calcium carbide: 1 kg of calcium carbide produces 0.41 kg of acetylene gas. Now, let's convert the mass of acetylene gas produced to volume using the ideal gas law at standard conditions (NTP - Normal Temperature and Pressure):

• Temperature (T) = 273.15 K
• Pressure (P) = 1 atm
• Volume (V) = 1 mole of any gas occupies 22.4 liters at standard temperature and pressure (STP)

Given that 1 mole of acetylene gas (𝐶2𝐻2C2​H2​) has a molar mass of approximately 26 g/mol, we can calculate the volume of acetylene gas produced from 1 kg of calcium carbide. Moles of 𝐶2𝐻2=Mass of 𝐶2𝐻2Molar mass of 𝐶2𝐻2Moles of C2​H2​=Molar mass of C2​H2​Mass of C2​H2​​ Moles of 𝐶2𝐻2=0.41 kg26 g/molMoles of C2​H2​=26g/mol0.41kg​ Now, we can find the volume of acetylene gas produced: Volume of 𝐶2𝐻2=Moles of 𝐶2𝐻2×Volume of 1 mole of gas at NTPVolume of C2​H2​=Moles of C2​H2​×Volume of 1 mole of gas at NTP Volume of 𝐶2𝐻2=Moles of 𝐶2𝐻2×22.4 liters/molVolume of C2​H2​=Moles of C2​H2​×22.4liters/mol Volume of 𝐶2𝐻2=(0.41 kg26 g/mol)×22.4 liters/molVolume of C2​H2​=(26g/mol0.41kg​)×22.4liters/mol Volume of 𝐶2𝐻2≈7.85 litersVolume of C2​H2​≈7.85liters Now, we need to determine how many hours of service can be derived from 7.85 liters of acetylene gas when burning 35 liters per hour. Hours of service=Volume of 𝐶2𝐻2Volume of gas consumed per hourHours of service=Volume of gas consumed per hourVolume of C2​H2​​ Hours of service=7.85 liters35 liters/hourHours of service=35liters/hour7.85liters​ Hours of service≈0.224 hoursHours of service≈0.224hours Since it's not practical to have a fraction of an hour of service, we need to round down to the nearest whole number, which is 0. Therefore, the correct answer is 0 hours of service.