C3H8(g) + 502(g) → 4H2O + 3CO2(g). Form the equation above the volume of oxygen at s.t.p. required to burn 50cm3 of propane is?
Question 1 Report
C3H8(g) + 502(g) → 4H2O + 3CO2(g). Form the equation above the volume of oxygen at s.t.p. required to burn 50cm3 of propane is?
Answer Details
C3H8 + 502 → 4H2O + 3CO2.
From the equation above 1 dm3 of C3H8 require 5dm3 of O2 at s.t.p.
∴ 50cm3 of C3H8 require x = (50) / (1000) x 5 x 1
x = (250) / (1000)
x = 250cm
From the equation above 1 dm3 of C3H8 require 5dm3 of O2 at s.t.p.
∴ 50cm3 of C3H8 require x = (50) / (1000) x 5 x 1
x = (250) / (1000)
x = 250cm
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