C₃H_8(g) + 50_2(g) → 4H₂O + 3CO_2(g). Form the equation above the volume of oxygen at s.t.p. required to burn 50cm³ of propane is?

250 cm³

150 cm³

100 cm ³

50 cm³

Correct answer is A

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

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C₃H_8(g) + 50_2(g) → 4H₂O + 3CO_2(g). Form the equation above the volume of oxygen at s.t.p. required to burn 50cm³ of propane is?