Answer:
hydrochloric acid
Explanation:
The hydrochloric acid in this reaction is the limiting reactant. A limiting reactant is the reactant that is used up in a chemical reaction. It determines the extent of the reaction.
Since the solution indicates a basic one after the end of the reaction, this suggests that more of the sodium hydroxide is still left unreacted with.
The reactant in excess supply here is the sodium hydroxide and the bulk of it is till left in solution.
Triprotic acid is a class of Arrhenius acids that are capable of donating three protons per molecule when dissociating in aqueous solutions. So the chemical reaction as described in the question, at the third equivalence point, can be show as: H3R + 3NaOH ⇒ Na3R + 3H2O, where R is the counter ion of the triprotic acid. Therefore, the ratio between the reacted acid and base at the third equivalence point is 1:3.
The moles of NaOH is 0.106M*0.0352L = 0.003731 mole. So the moles of H3R is 0.003731mole/3=0.001244mole.
The molar mass of the acid can be calculated: 0.307g/0.001244mole=247 g/mol.
The balanced chemical
reaction will be:
C4H8 + 6 O2 --> 4 CO2 + 4 H2O
We are given the amount of butene being combusted. This will be our
starting point.
136.6 g C4H8 (1 mol C4H8/ 56.11 g C4H8) (4 mol CO2/1 mol <span>C4H8</span>) ( 44.01 g CO2/ 1 mol CO2) = 428.6 g CO2
