Fuel gas usage problem (NTP basis):\nFrom 1 kg of calcium carbide (CaC2), about 0.41 kg of acetylene (C2H2) is produced with water. If an acetylene lamp burns 35 litres of gas per hour at NTP, how many hours of service are obtained from 1 kg CaC2?

Introduction / Context:
This stoichiometry and gas-law calculation links mass of generated acetylene to lamp runtime. At NTP (1 atm, 0°C), ideal-gas molar volume is approximately 22.414 L/mol. Converting produced mass to moles and then to volume allows a direct division by the consumption rate to obtain hours of operation.

Given Data / Assumptions:

• Mass of C2H2 produced from 1 kg CaC2: 0.41 kg = 410 g.
• Molar mass: M(C2H2) = 26 g/mol.
• NTP molar volume: ~22.414 L/mol.
• Lamp consumption: 35 L/h at NTP.
• Ideal-gas approximation; complete collection/usage of acetylene.

Concept / Approach:
Steps: convert 410 g C2H2 to moles; convert moles to litres at NTP; divide by the lamp’s volumetric burn rate to get time. Keep units consistent and use absolute NTP definition.

Step-by-Step Solution:
n_C2H2 = mass / molar mass = 410 g / 26 g/mol ≈ 15.769 mol.Volume at NTP = n * 22.414 L/mol ≈ 15.769 * 22.414 ≈ 353.2 L.Runtime = total volume / rate = 353.2 L / (35 L/h) ≈ 10.09 h.Rounded to the nearest option → 10 hours.

Verification / Alternative check:
Using 22.4 L/mol gives 15.769 * 22.4 ≈ 353.2 L, essentially identical; division still yields ≈10.1 h.

Why Other Options Are Wrong:
5 h is too low (would imply only ~175 L available); 15 h and 20 h require ~525–700 L of acetylene, far exceeding that produced; 22 h is even less plausible.

Common Pitfalls:
Using mass fractions directly instead of converting to moles; using STP/SATP values incorrectly; forgetting that NTP here is 0°C, 1 atm.

Final Answer:
10