Question- Calculate how many Glucose Liters will be produced if
there are 738 L of Carbon Dioxide
and 792 Liters of Water available?

123.2 L

Explanation:

We must first put down the required reaction equation;

6CO2(g) + 6H2O(g) > C6H12O6(g) + 6O2(g)

We know that 1 mol of a gas contains 22.4 L

hence the number of moles of carbon dioxide is obtained from;

1 mol = 22.4 L

x mol = 738 L

x = 738/22.4 = 32.9 mols

For water;

1 mol = 22.4 L

x mol = 792 L

x = 792/22.4

x = 35.4 mol

We obtain the limiting reactant;

6 mols of CO2 = 1 mole of glucose

32.9 mols of CO2 = 32.9 * 1/6 = 5.5 moles of glucose

For water;

6 mols of water yields 1 mol of glucose

35.4 mol of water yields 35.4 *  1/6 = 5.9 moles of glucose

CO2 is the limiting reactant

1 mol = 22.4 L

5.5 mol = 5.5 * 22.4 = 123.2 L