When the leaves of a tree were exposed to sunlight for a certain length of time 18.0 g of glucose (C₆H₁₂O₆) were made from carbon dioxide and water as a result of photosynthesis, and oxygen gas was also produced according to the following balanced equation - Leaving Cert Chemistry - Question c - 2020

To find the number of moles in 18.0 g of glucose, we can use the formula:

n = \frac{m}{M} = \frac{18.0}{180} = 0.1 ext{ moles of glucose}$$ Therefore, there are 0.1 moles of glucose in 18.0 g.

From the balanced equation, the ratio of moles of glucose (C₆H₁₂O₆) to moles of O₂ is:

$1 ext{ glc} : 6 ext{ O₂}$

This means for every 1 mole of glucose produced, 6 moles of oxygen are generated.

Since we have 0.1 moles of glucose:

Using the ratio discovered earlier: $0.1 ext{ moles of glucose} imes \frac{6 ext{ moles O₂}}{1 ext{ mole glucose}} = 0.6 ext{ moles of O₂}$

Thus, 0.6 moles of O₂ were produced from 18.0 g of glucose.

At standard temperature and pressure (s.t.p.), one mole of gas occupies 22.4 liters.

To find the volume occupied by 0.6 moles of O₂:

$0.6 ext{ moles O₂} imes 22.4 ext{ L/mole} = 13.44 ext{ L}$

Thus, the volume of O₂ produced is 13.44 liters.

Using the earlier ratio of C₆H₁₂O₆ to CO₂:

From the balanced equation, 6 moles of CO₂ are used to produce 1 mole of glucose.

Therefore, for 0.1 moles of glucose: $0.1 ext{ moles glucose} imes 6 = 0.6 ext{ moles CO₂}$

Now, calculating the mass of 0.6 moles of CO₂:

The molar mass of CO₂ is: $12 + (2 imes 16) = 44 ext{ g/mole}$

Thus, the mass of 0.6 moles of CO₂ is: $0.6 ext{ moles} imes 44 ext{ g/mole} = 26.4 ext{ g CO₂}$

So, 26.4 g of CO₂ was used up when 18.0 g of glucose were made.