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