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It's C guyzz !! It's really easy !! Just a bit tricky !!
Equation : 2C2H2 + 5O2 ----- 4CO2 + 2H2O
Volumes used : 20 cm3 + 500cm3 ------ ? + ?
Volumes actually
reacted : 20 cm3 + 50cm3 ---------- ? + ?
Volume of Oxygen reacted is 50 cm3 because according to the chemical equation , 2 of C2H2 requires 5 of O2 , so 20 of C2H2 requires 50 of O2 !!
NOW ACCORDING TO EQUATION :
1 cm3 + 2.5 cm3 ---------- 2 cm3 + 1 cm3
20cm3 + 50cm3 --------- 40cm3 + 20 cm3
Total volume of gases remaining : ( 500 - 50) + 40 = 490 cm3 !!!
The main part of the question to understand is that O2 was in excess !!!
Equation : 2C2H2 + 5O2 ----- 4CO2 + 2H2O
Volumes used : 20 cm3 + 500cm3 ------ ? + ?
Volumes actually
reacted : 20 cm3 + 50cm3 ---------- ? + ?
Volume of Oxygen reacted is 50 cm3 because according to the chemical equation , 2 of C2H2 requires 5 of O2 , so 20 of C2H2 requires 50 of O2 !!
NOW ACCORDING TO EQUATION :
1 cm3 + 2.5 cm3 ---------- 2 cm3 + 1 cm3
20cm3 + 50cm3 --------- 40cm3 + 20 cm3
Total volume of gases remaining : ( 500 - 50) + 40 = 490 cm3 !!!
The main part of the question to understand is that O2 was in excess !!!