Question

Given the following reaction: C3H8 + O2 ----> CO2 + H2Oa) If you start with14.8g of C3H8 and 3.44 g of O2 , determine the limiting reagent. b) determine the number of moles of carbon dioxide produced. c) determine the number of grams of H2O produced d) determine the number of grams of excess reagent left

Answer 1

a) Oxygen is the limiting reactant.

b) 0.0645 moles CO2= 2.84 grams CO2

c) Mass H2O = 1.55 grams

d) Mass propane remaining = 13.85 grams

Step-by-step explanation:

Step 1: Data given

MAss of propane (C3H8) = 14.8 grams

Mass of oxygen (O2) = 3.44 g/mol

Molar mass of O2 = 32.0 g/mol

Molar mass of propane = 44.1 g/mol

Step 2: The balanced equation

C3H8 + 5O2 ---->3CO2 + 4H2O

Step 3: Calculate moles

Moles =mass / molar mass

Moles propane = 14.8 grams / 44.1 g/mol

Moles propane = 0.3356 moles

Moles oxygen = 3.44 grams / 32.0 g/mol

Moles oxygen = 0.1075 moles

Step 4: Calculate the limiting reactant

For 1 mol C3H8 we need 5 moles O2 to produce 3 moles CO2 and 4 moles H2O

Oxygen is the limiting reactant. It will completely be consumed ( 0.1075 moles). Propane is in exess. There will react 0.1075/5 = 0.0215 moles

There will remain 0.3356 - 0.0215 = 0.3141 moles

step 5: Calculate mass of propane remaining

Mass propane = moles * molar mass

Mass propane remaining = 0.3141 moles * 44.1 g/mol

Mass propane remaining = 13.85 grams

Step 6: Calculate moles products

For 1 mol C3H8 we need 5 moles O2 to produce 3 moles CO2 and 4 moles H2O

For 0.1075 moles O2 we'll have 3/5 * 0.1075 = 0.0645 moles CO2

For 0.1075 moles O2 we'll have 4/5 * 0.1075 = 0.086 moles H2O

Step 7: calculate mass products

Mass = moles * molar mass

Mass CO2 = 0.0645 moles * 44.01 g/mol

Mass CO2 = 2.84 grams

Mass H2O = 0.086 grams * 18.02 g/mol

Mass H2O = 1.55 grams

Answer 2

a) O₂ is limiting reactant

b) 0.0645 moles of CO₂

c) 1.55g of H₂O

d) 13.85g C₃H₈

Step-by-step explanation:

The balanced reaction is:

a C₃H₈ + 5 O₂ → 3CO₂ + 4H₂O

1 mole of C₃H₈ reacts per 5 moles of O₂

14.8g of C₃H₈ and 3.44g of O₂ are:

14.8g × (1mol / 44.1g) = 0.3356moles C₃H₈

3.44g × (1mol / 32g) = 0.1075moles O₂

A complete reaction of 0.3356moles of C₃H₈ requires:

0.3356moles of C₃H₈ × (5mol O₂ / 1mol C₃H₈) = 1.678 mol O₂

As you start with just 0.1075 moles, O₂ is limiting reactant

b 0.1075moles O₂ produce:

0.1075moles O₂ × (3mol CO₂ / 5 mol O₂) = 0.0645 moles of CO₂

c Mass of H₂O produced from 0.1075moles O₂ is:

0.1075moles O₂ × (4mol H₂O / 5 mol O₂) × (18.01g / 1 mol H₂O) =

1.55g of H₂O

d 0.1075 moles O₂ react completely with:

0.1075 moles O₂ × (1mol C₃H₈ / 5mol O₂) = 0.0215mol C₃H₈.

As you have 0.3356 moles C₃H₈, moles that don't react are:

0.3356moles C₃H₈ - 0.0215mol C₃H₈ = 0.3141 moles C₃H₈. In grams:

0.3141 moles C₃H₈ (44.1g / 1mol) = 13.85g C₃H₈