Answer to Question #175698 in Chemistry for beeta

Question

C3H8 combusts. molar masses:
tricarbon octahydride: 44.11 g/moloxygen gas: 32 g/molwater: 18.02 g/molcarbon dioxide: 44.01 g/mol
a. Write a balanced chemical reaction.
b. If 5.34 g of tricarbon octahydride reacts with 25.2 g of oxygen gas, what is the limiting reactant? (tricarbon octahydride)
c. How many grams of water is produced? (8.72168)
d. How much excess reactant is left over after the reaction is complete? (5.84)
e. What is the percent yield of water if 6.98 g of water is actually produced in a laboratory experiment? (80.0459)

Answer 1

Solution:

(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%.