You preform an experiment using 4.0 m hcl. And 2.4 m naoh. The experiment calls for 5.0 ml of hcl and 10.0 ml of naoh. How many moles of nacl are produced from this reaction.

Respuesta :

Answer:

0,020 moles of NaCl

Explanation:

For the reaction:

NaOH + HCl → NaCl + H₂O

The moles of NaOH are:

10,0 mL ≡ 0,0100L × 2,4mol/L = 0,024 moles of NaOH.

In the same way, moles of HCl are:

5,0 mL ≡ 0,0050L × 4,0mol/L = 0,020 moles of HCl.

The limiting reactant is the reactant that is completely used in the reaction.

In this case, as moles of HCl are less than moles of NaOH, limiting reactant is HCl.

That means that moles of NaCl that are produced are the same moles of HCl, 0,020 moles

I hope it helps!

0.020 moles of NaCl is produced in this reaction.

The equation of the reaction is; HCl(aq) + NaOH(aq) -----> NaCl(aq) + H2O(l)

Number of moles of NaOH = 2.4 M × 10/1000 L = 0.024 moles

Number of moles of HCl = 4.0 M × 5/1000 L = 0.020 moles

Since the reaction is 1:1 then HCl is the limiting reactant.

Therefore, 0.020 moles of NaCl is produced

Mass of NaCl produced = mass × molar mass

Molar mass of NaCl = 58.5 g/mol × 0.020 moles  = 1.17 g

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