Consider the combustion of octane: 2C8H18(g) + 2502(g) ===> 16CO2(g) + 18H2O(g)
Calculate the mass of water (H20) produced from 11.2 grams of octane (C3H18) assuming excess
oxygen.
15.9 g
O 0.196 g
O 4.31 g
1.77g
14.1g

15.9 g

Explanation:

(Take the atomic mass of C=12.0, H=1.0, O=16.0)

no. of moles = mass / molar mass

no. of moles of octane used = 11.2 / (12.0x8 + 1x18)

= 0.0982456 mol

Since oxygen is in excess and octane is the limiting reagent, the no. of moles of H2O depends on the no. of moles of octane used.

From the balanced equation, the mole ratio of octane : water = 2:18 = 1: 9,

so this means, one mole of octane produced 9 moles of water.

Using this ratio, we can deduce that (y is the no. of moles of water produced):

y = 0.0982456x9

y= 0.88421 mol

Since mass = no. of moles x molar mass,

mass of water produced = 0.88421  x (1.0x2+16.0)

=15.9 g

it should be all of them because it asks if you included all of those in your answer

explanation:

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