# Ideal gas law and percent yield questionIn a calibration experiment, a 1.01 millimole sample of Na2CO3 gave 0.85 millimoles of CO2 gas. If a 0.376 g of pure Na2CO3 was reacted with excess acid,...

Ideal gas law and percent yield question

In a calibration experiment, a 1.01 millimole sample of Na2CO3 gave 0.85 millimoles of CO2 gas. If a 0.376 g of pure Na2CO3 was reacted with excess acid, what volume of gas will be measured on this apparatus (in Litres at STP)?

ndnordic | Certified Educator

In the calibration you measured a yield of 84.158 % (.85/1.01*100) for the amount of gas produced by the decomposition of sodium carbonate. Note that this assumes a pure sample of sodium carbonate was used.

In the actual experiment 0.376 g of sodium carbonate was the starting amount which is 0.376g/105.987 g/mole = 3.5476 milimoles of sodium carbonate. Since the equipment is finding 84.158% of the CO2 produced, there would be 3.5476 * 0.84158 = 2.9856 milimoles of CO2 measured.  Based on 22.4 mL/ milimole for CO2 at STP, you would then measure 2.9856*22.4 = 66.877 mL or 0.0669 L of CO2