Respuesta :
ANSWER
The volume of CO2 used at STP is 77.325L
EXPLANATION
Given that;
The mass of C6H12O6 is 622g
[tex]\text{ 6CO}_2\text{ + 6H}_2O\text{ }\rightarrow\text{ C}_6H_{12}O_6\text{ + 6O}_2[/tex]Follow the steps below to find the volume of CO2 reacted
Step 1; Find the number of moles of C6H12O6 using the below formula
[tex]\text{ mole = }\frac{\text{ mass}}{\text{ molar mass}}[/tex]Recall, that the molar mass of C6H12O6 is 180.16 g/mol
[tex]\begin{gathered} \text{ mole = }\frac{\text{ 622}}{\text{ 180.16}} \\ \text{ mole = 3.452 moles} \end{gathered}[/tex]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
[tex]\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{ }\times\text{ xL = 3.452 mol CO}_2\text{ }\times\text{ 22.4 L} \\ \text{ Isolate xL} \\ \text{ xL CO}_2\text{ = }\frac{3.452\cancel{mol}CO_2\text{ }\times22.4\text{ L}}{1\cancel{mol}\text{ CO}_2} \\ \text{ xL = 3.452 }\times\text{ 22.4} \\ \text{ xL = 77.325 L} \end{gathered}[/tex]
Therefore, the volume of CO2 used at STP is 77.325L
