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If in Part II, you mixed (carefully measured) 25.0 mL of 0.81 M NaOH with 65.0 mL of 0.33 M HCl, which of the two reagents is the limiting reagent for heat of reaction

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Sebassandinidn

Answer:

NaOH is the limiting reactant.

Explanation:

Hello there!

In this case, according to the given information, it turns out firstly necessary to write out the chemical reaction between NaOH and HCl:

[tex]NaOH+HCl\rightarrow NaCl+H_2O[/tex]

Thus, since they react in a 1:1 mole ratio; we can now calculate the moles of each substance by using their volumes and molarities:

[tex]n_{NaOH}=0.0250L*0.81mol/L=0.02025molNaOH\\\\n_{HCl}=0.0650L*0.33mol/L=0.02145molHCl[/tex]

Now, since NaOH is in a fewer proportion, we infer just 0.02025 moles of HCl are consumed so that 0.0012 moles of this acid remain unreacted; in such a way, we infer that the NaOH is the limiting reactant for this reaction.

Regards!

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