Question

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

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