Question

A neutralization reaction was performed using NaOH (volume - 250 ml., concentration - 0.500 M) and nitric acid (volume - 25.0 ml., concentration - 0.550 M),
Calculate the moles of the limiting reactant.
NaOH + HNO, - NaNO, +H2O
Give your answer to three significant figures.
mol

Answer 1

0.01375 moles of nitric acid (HNO₃) which is the limiting reactant

Explanation:

We have the following chemical reaction:

NaOH + HNO₃ → NaNO₃ + H₂O

Now to determine the number of moles of each reactant we use the following formula:

molar concentration = number of moles / volume (L)

number of moles = molar concentration × volume (L)

number of moles of NaOH = 0.5 × 0.250 = 0.125 moles

number of moles of HNO₃ = 0.55 × 0.025 = 0.01375 moles

From the chemical reaction we see that 1 mole of NaOH reacts with 1 mole of HNO₃, so 0.125 moles of NaOH will react with 0.125 moles of HNO₃ but we only have 0.01375 moles of HNO₃ available, so the limiting reactant is HNO₃.

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