Question

During photosynthesis, a potato plant undergoes this reaction:6CO2 + 6H₂O --> C6H12O6 + 602Calculate the litres of CO2 at STP that are required to produce 622 g of C6H12O6Show your work

Answer 1

The volume of CO2 used at STP is 77.325L

Step-by-step explanation

Given that;

The mass of C6H12O6 is 622g

`\text{ 6CO}_2\text{ + 6H}_2O\text{ }\rightarrow\text{ C}_6H_(12)O_6\text{ + 6O}_2`

Follow the steps below to find the volume of CO2 reacted

Step 1; Find the number of moles of C6H12O6 using the below formula

`\text{ mole = }\frac{\text{ mass}}{\text{ molar mass}}`

Recall, that the molar mass of C6H12O6 is 180.16 g/mol

`\begin{gathered} \text{ mole = }\frac{\text{ 622}}{\text{ 180.16}} \\ \text{ mole = 3.452 moles} \end{gathered}`

The moles of C6H12O6 is 3.452 moles

Step 2; Find the number of moles of CO2 using a stoichiometery ratio

In the above equation, 1 mol CO2 reacted to give 1 mol C6H12O6

Since 1 mol of CO2 is equivalent to 1 mole of C6H12O6

Therefore, the number of moles of CO2 is 3.452 moles

Step 3; Find the volume of CO2 at STP

Recall, 1 mole of a gas is equivalent to 22.4 L/mol

`\begin{gathered} \text{ 1 mol CO}_2\text{ }\rightarrow\text{ 22.4 L/mol} \\ \text{ 3.452 mol CO}_2\text{ }\rightarrow\text{ xL CO}_2 \\ \text{ cross multiply} \\ \text{ 1 mol CO}_{2\text{ }}\text{ }*\text{ xL = 3.452 mol CO}_2\text{ }*\text{ 22.4 L} \\ \text{ Isolate xL} \\ \text{ xL CO}_2\text{ = }\frac{3.452\cancel{mol}CO_2\text{ }*22.4\text{ L}}{1\cancel{mol}\text{ CO}_2} \\ \text{ xL = 3.452 }*\text{ 22.4} \\ \text{ xL = 77.325 L} \end{gathered}`

Therefore, the volume of CO2 used at STP is 77.325L