Question

Asked Oct 26, 2019

LP gas burns according to the exothermic reaction:

C3H8(g)+5 O2(g)→3 CO2(g)+4 H2O(g)ΔH∘rxn=−2044 kJ

What mass of LP gas is necessary to heat 1.5 L of water from room temperature (25.0 °C) to boiling (100.0 °C)? Assume that during heating, 15% of the heat emitted by the LP gas combustion goes to heat the water. The rest is lost as heat to the surroundings.

Step 1

**Given:**

The balanced chemical equation:

C_{3}H_{8}(g)+5 O_{2}(g)→3 CO_{2}(g)+4 H_{2}O(g)

ΔH^{∘}_{rxn}=−2044 kJ

Volume of water = 1.5 L

Heat emitted = 15%

Change in T = (25.0 °C) to (100.0 °C)

Step 2

1.5L means 1500 ml since density of water = 1,

1 ml = 1 g

1500 ml = 1500 gr

The heat absorbed by water (Q) = m x c x delta T

m = mass of the water = 1500 gr

c = specific heat of the water = 4.184 J/g ºC

delta...

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