How many grams of H2O are formed from 0.36 mol of C4H10?
g H2O
The balanced chemical equation for the combustion of C4H10 (butane) is:
C4H10 + 6.5O2 -> 4CO2 + 5H2O
From the equation, we can see that 4 moles of H2O are formed for every 1 mole of C4H10 that reacts.
Given that there are 10 moles of H2O produced for every 4 moles of C4H10, we can calculate the number of moles of H2O produced from 0.36 mol of C4H10:
0.36 mol C4H10 x (10 mol H2O / 4 mol C4H10) = 0.9 mol H2O
Now we can convert this into grams by multiplying by the molar mass of water (H2O):
0.9 mol H2O x 18.015 g/mol = 16.21 g H2O
Therefore, 16.21 grams of H2O are formed from 0.36 mol of C4H10.