Question

Suppose you have 5.1 g of ethane mixed with

31.4 g of oxygen. What is the limiting reactant?

a. Ethane
b. Oxygen
c. Both react completely
d. Not enough information

Answer 1

The limiting reactant between ethane and oxygen is determined through stoichiometry using the balanced combustion equation. Ethane is the limiting reactant due to having fewer moles than needed to fully react with the available oxygen. Oxygen is in excess by 12.32 grams.

Step-by-step explanation:

To determine the limiting reactant between ethane (C2H6) and oxygen (O2), we need to consider the balanced chemical equation for the combustion of ethane, which is:

2C2H6(g) + 7O2(g) → 4CO2(g) + 6H2O(l)

First, we calculate the moles of each reactant:

• Moles of C2H6 = 5.1 g / 30.07 g/mol = 0.17 mol
• Moles of O2 = 31.4 g / 32.00 g/mol = 0.98 mol

According to the equation, 1 mole of C2H6 requires 3.5 moles of O2. For 0.17 moles of C2H6, we need 0.17 mol * 3.5 mol O2/mol C2H6 = 0.595 mol of O2. Since we have more moles of O2 (0.98 mol) than needed (0.595 mol), ethane is the limiting reactant.

Now let's calculate the excess of O2:

Excess O2 = Initial moles of O2 - Moles of O2 required = 0.98 mol - 0.595 mol = 0.385 mol

The amount in grams of excess O2 is: 0.385 mol * 32.00 g/mol = 12.32 g.

Therefore, the limiting reactant is ethane (C2H6), and there is 12.32 g of oxygen in excess.