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

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

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

the correct answer would be d. hope this !