The catabolism of glucose has an equation of C6H12O6 + 6O2 = 6CO2 +6 H20. Hence for every mole of glucose, 6 moles of CO2 is produced. Given 22 grams of CO2, that is 0.5 mol CO2, we multiply this by 1/6, we get the number of moles of glucose equal to 1/12 mol. The mass of glucose needed is obtained by multiplying this by molar mass of glucose which is 180 g/mol. This is equivalent 15 grams of glucose.
The answer would be C) Observe and ask questions because if we do process of elimination we see that drawing conclusions is NOT the first step, forming a hypothesis and illustrating the process is part of the method, but NOT the first step.
Answer:
74mL
Explanation:
Given parameters:
Molar mass of citric acid = 192g/mol
Molar mass of baking soda = 84g/mol
Concentration of citric acid = 0.8M
Mass of baking powder = 15g
Unknown parameters:
Volume of citric acid = ?
Solution
Equation of the reaction:
C₆H₈O₇ + 3NaHCO₃ → Na₃C₆H₅O₇ + 3H₂O + 3CO₂
Procedure:
- We work from the known parameters to the unknown. From the statement of the problem, we can approach the solution from the parameters of the baking powder.
- From the baking powder, we can establish a molar relationship between the two reactants. We employ the mole concept in this regard.
- We find the number of moles of the baking powder that went into the reaction using the expression below:
Number of moles = 
Number of moles =
= 0.179mole
- From the equation of the reaction, we can find the number of moles of the citric acid:
3 moles of baking powder reacted with 1 mole of citric acid
0.179 moles of baking powder would react with
:
This yields 0.059mole of citric acid
- To find the volume of the citric acid, we use the mole expression below:
Volume of citric acid = 
Volume of citric acid =
= 0.074L
Expressing in mL gives 74mL