What is the amount of oxygen (in g) required for the complete combustion of 900 g of glucose?

Steps to Solve

1. Write the balanced chemical equation

The balanced equation for the complete combustion of glucose (C₆H₁₂O₆) is:

$$ C_6H_{12}O_6 + 6 O_2 \rightarrow 6 CO_2 + 6 H_2O $$

This indicates that one mole of glucose reacts with six moles of oxygen.

2. Calculate the moles of glucose

First, we need to find the number of moles of glucose in 900 g. The formula for moles is:

$$ \text{Moles} = \frac{\text{mass (g)}}{\text{molar mass (g mol}^{-1}\text{)}} $$

For glucose:

$$ \text{Moles of glucose} = \frac{900 , \text{g}}{180 , \text{g mol}^{-1}} = 5 , \text{mol} $$

3. Calculate the required moles of oxygen

Using the balanced equation, we know that 1 mole of glucose requires 6 moles of oxygen. Therefore, for 5 moles of glucose:

$$ \text{Moles of oxygen} = 5 , \text{mol glucose} \times 6 , \frac{\text{mol O}_2}{\text{mol glucose}} = 30 , \text{mol O}_2 $$

4. Calculate the mass of oxygen needed

Now, using the molar mass of oxygen (32 g mol⁻¹), we can find the mass required:

$$ \text{Mass of oxygen} = 30 , \text{mol O}_2 \times 32 , \text{g mol}^{-1} = 960 , \text{g} $$