Question

If 37.8 grams of pentane (C5H12) are burned in excess oxygen, how many grams of H2O...

If 37.8 grams of pentane (C5H12) are burned in excess oxygen, how many grams of H2O will be produced?

Homework Answers

Answer #1

The balanced equation for combustion of pentane is

C5H12 + 8 O2 ------> 6 H2O + 5 CO2

number of moles of pentane = 37.8 g / 72.15 g/mol = 0.524 mole

from the balanced equation we can say that

1 mole of C5H12 produces 6 mole of H2O so

0.524 mole of C5H12 will produce

= 0.524 mole of C5H12 *(6 mole of H2O / 1 mole of C5H12)

= 3.14 mole of H2O

Mass of 1 mole of H2O = 18.015 g

so the mass of 3.14 mole of H2O = 3.14 * 18.015 = 56.6 g

Therefore, the mass of H2O produced would be 56.6 g

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