Consider the following combustion reaction: C3H8(g)+5O2(g)→3CO2(g)+4H2O(g) At STP (atm), when 64.4 L of propane is reacted with 87.0 L of oxygen gas, volume of the CO2 gas produced is:
A
105 L of CO2
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
258 L of CO2
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
39.2 L of CO2
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
52.2 L of CO2
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
Open in App
Solution
The correct option is D 52.2 L of CO2 1 mole of propane requires 5 moles of oxygen 22.4 L of propane requires (5×22.4) L of oxygen 64.4 L of propane will require 5×22.422.4×64.4=322 L of oxygen Therefore, oxygen is the limiting reagent. 5 moles of oxygen produce 3 moles of CO2 (5×22.4) L of oxygen produces (3×22.4) L of CO2 So, 87 L of oxygen will produce 3×22.45×22.4×87= 52.2 L of CO2