To determine which reactant is limiting and which is in excess, we need to compare the number of moles of each reactant to the stoichiometric ratio in the balanced equation.
First, we need to convert the mass of each reactant to moles using the molar mass:
Molar mass of propane (C3H8) = 3(12.01 g/mol) + 8(1.01 g/mol) = 44.11 g/mol
Number of moles of propane = 54.7 g / 44.11 g/mol = 1.24 mol
Molar mass of oxygen (O2) = 2(16.00 g/mol) = 32.00 g/mol
Number of moles of oxygen = 89.6 g / 32.00 g/mol = 2.80 mol
Next, we compare the moles of each reactant to the stoichiometric ratio in the balanced equation:
For propane (C3H8):
- From the balanced equation, we see that 1 mole of propane (C3H8) reacts with 5 moles of oxygen (O2).
- The number of moles of oxygen available is 2.80 mol, which means it could react with:
2.80 mol / 5 mol = 0.56 mol of propane
For oxygen (O2):
- From the balanced equation, we see that 1 mole of propane (C3H8) reacts with 5 moles of oxygen (O2).
- The number of moles of propane available is 1.24 mol, which means it could react with:
1.24 mol * 5 mol = 6.20 mol of oxygen
Since we have 0.56 mol of propane available but only 1.24 mol of propane, propane is the limiting reactant. Oxygen is in excess.
Therefore, the limiting reactant is propane (C3H8) and the reactant in excess is oxygen (O2).
10. If 54.7 grams of propane (C3H8) and 89.6 grams of oxygen (O2) are available in the balanced combustion reaction below:
C_3 H_8+5O_2⟶3CO_2+4H_2 O
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