A sample of gaseous hydrocarbon occupying 1.12 litre at NTP, when completely burnt in air produced 2.2 g CO2 and 1.8 g H2O. Calculate the mass of hydrocarbon taken and the volume of O2 at NTP required for its combustion.
2.24 L
You know that a hydrocarbon is made up of C and H
∴ CaHb + (a + b4)O2 → a CO2 + b2 H2O (I) is the combustion of the hydrocarbon.
1 mole of CaHb = (44a) g of CO2
1.12 L of CaHb = (1.12×44a22.4) g of CO2
From question 1.12×44a22.4 = 2.2g ⇒ a = 1
Again, 22.4 L of CaHb = (b2 × 18)g of H2O
1.12 of CaHb = 1.12×b2×1822.4
From question, 1.12×b2×1822.4 = 1.8 ⇒ b = 4
∴ The hydrocarbon of methane (CH4)
∴ Number of moles of oxygen = ( 1 + 44) = 2 moles.
From reaction,
22.4 L of CH4 = 2 × 22.4 L of O2
1.12 L of CH4 = 1.12×2×22.422.4
1.12 L of CH4 = 2.24L of O2