The volume in litters of CO2 liberated at STP when 10 grams of 90% pure limestone is heated completely is

Explanation

The correct answer is d. 2.016 L.Calculation:Calculate the amount of pure limestone in the sample:○ Given purity = 90%○ Total mass = 10 g○ Pure limestone mass = purity * total mass = 0.9 * 10 g = 9 gConvert the mass of pure limestone to moles:○ Molar mass of CaCO3 (limestone) = 100.09 g/mol○ Moles of CaCO3 = mass / molar mass = 9 g / 100.09 g/mol = 0.09 molesDetermine the moles of CO2 produced:○ From the balanced chemical equation for limestone decomposition: CaCO3 (limestone) -> CaO + CO2○ 1 mole of CaCO3 decomposes to produce 1 mole of CO2.○ Therefore, moles of CO2 produced = moles of CaCO3 = 0.09 molesCalculate the volume of CO2 at STP:○ Standard temperature and pressure (STP) = 0 °C and 1 atm○ Ideal gas law: PV = nRT○ Assuming ideal behavior for CO2 at STP: P = 1 atm, V = unknown, n = 0.09 moles, R = 0.0821 L atm/mol K, T = 273.15 K (0 °C)○ Solve for V: V = nRT/P = 0.09 moles * 0.0821 L atm/mol K * 273.15 K / 1 atm = 2.016 LTherefore, 2.016 liters of CO2 are liberated at STP when 10 grams of 90% pure limestone is heated completely.