The balanced equation for the above reaction is as follows
C₆H₁₂O₆(s) + 6O₂(g) --> 6H₂O(g) + 6CO₂<span>(g)
the limiting reactant in the equation is glucose as the whole amount of glucose is used up in the reaction.
the amount of </span>C₆H₁₂O₆ used up - 13.2 g
the number of moles reacted - 13.2 g/ 180 g/mol = 0.073 mol
stoichiometry of glucose to CO₂ - 1:6
then number of CO₂ moles are - 0.073 mol x 6 = 0.44 mol
As mentioned this reaction takes place at standard temperature and pressure conditions,
At STP 1 mol of any gas occupies 22.4 L
Therefore 0.44 mol of CO₂ occupies 22.4 L/mol x 0.44 mol = 9.8 rounded off - 10.0 L
Answer is B) 10.0 L CO₂
Option B is the correct answer
Answer:
2.16 × 10⁻³
Explanation:
Step 1: Given data
Concentration of the acid (Ca): 0.260 M
Acid dissociation constant (Ka): 1.80 × 10⁻⁵
Step 2: Write the acid dissociation equation
HC₂H₃O₂(aq) + H₂O(l) ⇄ C₂H₃O₂⁻(aq) + H₃O⁺(aq)
Step 3: Calculate the concentration of H₃O⁺ at equilibrium
We will use the following expression.
![[H_3O^{+} ]= \sqrt{Ka \times Ca } = \sqrt{1.80 \times 10^{-5} \times 0.260 } = 2.16 \times 10^{-3}](https://tex.z-dn.net/?f=%5BH_3O%5E%7B%2B%7D%20%5D%3D%20%5Csqrt%7BKa%20%5Ctimes%20Ca%20%7D%20%3D%20%5Csqrt%7B1.80%20%5Ctimes%2010%5E%7B-5%7D%20%5Ctimes%200.260%20%7D%20%3D%202.16%20%5Ctimes%2010%5E%7B-3%7D)