The balanced equation for the combustion of octane is:
(2C_8H_{18} + 25O_2 \rightarrow 16CO_2 + 18H_2O)
How many grams of CO2 are formed from 8.7 mol of C8H18?
How many grams of (H_2O) are formed from 0.86 mol of (C_8H_{18})?”
To find the number of grams of CO2 formed from 8.7 mol of C8H18, we first need to use the balanced equation to determine the mole ratio between C8H18 and CO2. From the balanced equation, we can see that 2 moles of C8H18 produces 16 moles of CO2.
1. Convert 8.7 mol of C8H18 to moles of CO2:
8.7 mol C8H18 x (16 mol CO2 / 2 mol C8H18) = 69.6 mol CO2
2. Convert moles of CO2 to grams using the molar mass of CO2:
69.6 mol CO2 x (44.01 g/mol CO2) = 3066.96 g CO2
Therefore, 3066.96 grams of CO2 are formed from 8.7 mol of C8H18.
To find the number of grams of H2O formed from 0.86 mol of C8H18, we follow a similar process:
1. Convert 0.86 mol of C8H18 to moles of H2O:
0.86 mol C8H18 x (18 mol H2O / 2 mol C8H18) = 7.74 mol H2O
2. Convert moles of H2O to grams using the molar mass of H2O:
7.74 mol H2O x (18.02 g/mol H2O) = 139.37 g H2O
Therefore, 139.37 grams of H2O are formed from 0.86 mol of C8H18.