Question

1) 2 C3H7OH + 9 O2 ⇒ 6 CO2 - 8 H2O

If 8 grams of C3H7OH react with 4 grams of O₂

a) What is the limiting reactant? b) What is the excess reactant? c) How many moles of CO₂ can be produced? d) How many grams of CO₂ can be produced? e) How many grams of excess reactant react? f) How many grams of excess reactant remain?​

Answer 1

a) To determine the limiting reactant, we need to compare the amount of product that can be produced by each reactant. To do this, we need to convert the mass of each reactant to moles.

For C3H7OH:

moles = mass / molar mass = 8 g / 60.1 g/mol = 0.133 mol

For O2:

moles = mass / molar mass = 4 g / 32 g/mol = 0.125 mol

According to the balanced equation, 2 moles of C3H7OH react with 9 moles of O2 to produce 6 moles of CO2. Therefore, for 0.133 mol of C3H7OH, we need:

moles of O2 = (9/2) x 0.133 mol = 0.597 mol

Since we only have 0.125 mol of O2, it is the limiting reactant. Therefore, O2 is the limiting reactant.

b) C3H7OH is the excess reactant.

c) Using the balanced equation, we can see that 2 moles of C3H7OH react with 9 moles of O2 to produce 6 moles of CO2. Therefore, for 0.125 mol of O2, we can produce:

moles of CO2 = (6/9) x 0.125 mol = 0.083 mol

d) The molar mass of CO2 is 44.01 g/mol. Therefore, the mass of CO2 produced is:

mass of CO2 = moles of CO2 x molar mass of CO2

mass of CO2 = 0.083 mol x 44.01 g/mol = 3.65 g

e) To calculate how much excess reactant is used, we need to first determine how much of the excess reactant is present:

moles of C3H7OH = mass / molar mass = 8 g / 60.1 g/mol = 0.133 mol

moles of C3H7OH needed = (2/9) x moles of O2 = (2/9) x 0.125 mol = 0.028 mol

mol of excess C3H7OH = moles of C3H7OH present - moles of C3H7OH needed

mol of excess C3H7OH = 0.133 mol - 0.028 mol = 0.105 mol

moles of O2 needed = (9/2) x mol of excess C3H7OH = (9/2) x 0.105 mol = 0.473 mol

moles of O2 used = moles of limiting reactant = 0.125 mol

moles of excess C3H7OH used = (1/2) x moles of O2 used = (1/2) x 0.125 mol = 0.0625 mol

f) To calculate how much excess reactant remains, we need to subtract the moles of excess reactant used from the initial moles of excess reactant:

mol of excess C3H7OH remaining = 0.105 mol - 0.0625 mol = 0.0425 mol

The mass of excess reactant remaining can be calculated as follows:

mass of excess C3H7OH remaining = mol of excess C3H7OH remaining x molar mass of C3H7OH

mass of excess C3H7OH remaining = 0.0425 mol x 60.1 g/mol = 2.55