One litre of oxygen at STP is made to react with three litres of carbon monoxide at STP. Calculate the mass of each substance found after the reaction. Which one is the limiting reactant?

One litre of oxygen at STP is made to react with three litres of carbo

1.	One litre of oxygen at STP is made to react with three litres of carbon monoxide at STP. Calculate the mass of each substance found after the reaction. Which one is the limiting reactant?
Answer» Solution :The chemical equation representing the reaction is : `UNDERSET("2 vol.")(2CO)+underset("1 vol.")(O_(2))RARR underset("2 vol.")(2CO_(2))` Step 1. To calculate the VOLUME and mass of CO (at STP) left unused after the reaction. Applying Gay Lussac's Law of Gaseous Volumes, 1 Vol. of `O_(2)` reacts with CO = 2 Vol. `therefore"1 litre of "O_(2)" will react with CO = 2 litres at STP"` `"But the volume of CO taken = 3 litre at STP"` `therefore"Volume of CO (at STP) left unused "=3-2-1" litre."` Now, by mole concept, mass of 22.4 litres of CO at STP = Molecular mass is grams `=12+16=28g` `therefore"Mass of 1 litre of CO at STP"=(28)/(22.4)xx1=1.25g` Thus, the mass of CO left unused = 1.25g Step 2. To calculate the volume and mass of `CO_(2)` formed from 1 litre of `O_(2)` at sTP. Applying Gay Lussac's Law of Gaseous Volumes, 1 Vol. of `O_(2)` produces `CO_(2)=" 2 Vol."` `therefore"1 litre of "O_(2)" will produce CO"_(2)=" 2 litres at sTP"` By mole concept, Mass of 22.4 litres of `CO_(2)` at STP = Molecular mass in grams `=12+2xx16=44g` `therefore"Mass of 2 litres of "CO_(2)" at STP"=(44)/(22.4)xx2=3.928g` Thus, the mass of `CO_(2)` produced = 3.928g. Step 3. As oxygen has been COMPLETELY used up, hence oxygen is the limiting REACTANT.