Question

ii. For the combustion of sucrose: C12H22011 + 1202 12CO2 + 11H2O
There are 10.0 g of sucrose and 10.0 g of oxygen reacting. Which is the limiting reagent and
calculate the amount f formation of carbon dioxide?​

Answer 1

Answer:

Problem #1: For the combustion of sucrose:

C12H22O11 + 12O2 ---> 12CO2 + 11H2O

there are 10.0 g of sucrose and 10.0 g of oxygen reacting. Which is the limiting reagent?

Solution path #1:

1) Calculate moles of sucrose:

10.0 g / 342.2948 g/mol = 0.0292146 mol

2) Calculate moles of oxygen required to react with moles of sucrose:

From the coefficients, we see that 12 moles of oxygen are require for every one mole of sucrose. Therefore:

0.0292146 mol times 12 = 0.3505752 mole of oxygen required

3) Determine limiting reagent:

Oxygen on hand ⇒ 10.0 g / 31.9988 g/mol = 0.3125 mol

Since the oxygen required is greater than that on hand, it will run out before the sucrose. Oxygen is the limiting reagent.

Solution path #2:

1) Calculate moles:

sucrose ⇒ 0.0292146 mol

oxygen ⇒ 0.3125 mol

2) Divide by coefficients of balanced equation:

sucrose ⇒ 0.0292146 mol / 1 mol = 0.0292146

oxygen ⇒ 0.3125 mol / 12 mol = 0.02604

Oxygen is the lower value. It is the limiting reagent.

The second method above will be the preferred method to determine the limiting reagent in the following problems.

Problem #2: Calculate the number of NaBr formula units formed when 50 NBr3 molecules and 57 NaOH formula units react?

2NBr3 + 3NaOH ---> N2 + 3NaBr + 3HOBr