Question

C3H8 combusts.

Answer 1

Answer:

The balanced chemical equation:

C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(l)

Explanation:

(a): The balanced chemical equation:

C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(l)

(b):  Determine moles of each reactant:

(5.34 g C3H8) × (1 mol C3H8 / 44.11 g C3H8) = 0.1211 mol C3H8

(25.2 g O2) × (1 mol O2 / 32.00 g O2) = 0.7875 mol O2

According to the chemical equation above: n(C3H8) = n(O2)/5

Choose one reactant and determine how many moles of the other reactant are necessary to completely react with it. Let's choose C3H8:

n(O2) = 5 × n(C3H8) = 5 × 0.1211 mol = 0.6055 mol

The calculation above means that we need 0.6055 mol of O2 to completely react with 0.1211 mol C3H8.

We have 0.7875 mol O2 and therefore more than enough oxygen.

Thus oxygen (O2) is in excess and tricarbon octahydride (C3H8) must be the limiting reactant.

The limiting reactant is tricarbon octahydride (C3H8).

(c):

(5.34 g C3H8) × (1 mol C3H8 / 44.11 g C3H8) × (4 mol H2O/ 1 mol C3H8) × (18.02 g H2O / 1 mol H2O) = 8.726 g H2O

8.726 grams of water (H2O) is produced.

(d):

0.7875 mol O2 - 0.6055 mol of O2 = 0.182 mol O2 (excess O2)

(0.182 mol O2) × (1 mol O2 / 32.00 g O2) = 5.824 g O2

5.824 grams of oxygen gas (O2) is left over after the reaction is complete.

(e):

%H2O = (6.98 g / 8.726 g) × 100% = 79.99% = 80.00%

The percent yield of water (H2O) is 80.00%.