How many grams of H2O are formed from 0.48 mol of C8H18?
To determine the amount of water (H2O) formed from 0.48 mol of C8H18, we can start by writing a balanced chemical equation for the combustion of C8H18:
C8H18 + 12.5O2 → 8CO2 + 9H2O
From the balanced equation, we see that for every mole of C8H18 combusted, 9 moles of H2O are formed.
Therefore, if 0.48 mol of C8H18 is combusted, the amount of H2O formed is:
0.48 mol C8H18 * 9 mol H2O/1 mol C8H18 = 4.32 mol H2O
Now, we need to convert the moles of H2O to grams. The molar mass of H2O is approximately 18.01528 g/mol.
4.32 mol H2O * 18.01528 g/mol = 77.86 g H2O
Therefore, 77.86 grams of H2O are formed from 0.48 mol of C8H18.