Answered

Find the best answers to your questions at Westonci.ca, where experts and enthusiasts provide accurate, reliable information. Explore comprehensive solutions to your questions from knowledgeable professionals across various fields on our platform. Explore comprehensive solutions to your questions from a wide range of professionals on our user-friendly platform.

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

Sagot :

sebassandin

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!

Thanks for stopping by. We strive to provide the best answers for all your questions. See you again soon. Thank you for visiting. Our goal is to provide the most accurate answers for all your informational needs. Come back soon. Get the answers you need at Westonci.ca. Stay informed by returning for our latest expert advice.