Consider the neutralization between the following two solutions:
100.0 mL 0.9285 M hydrochloric acid, HCl
150.0 mL 1.8269 M sodium hydroxide, NaOH
How many moles of water are produced by this neutralization?
HCl + NaOH ==> NaCl + H2O
100.0 mL 0.9285 M hydrochloric acid, HCl
150.0 mL 1.8269 M sodium hydroxide, NaOH
mols HCl initially = mL/1000 x M = 0.09285
mols NaOH initially = 0.2740
mols H2O produced IF we use HCl as the limiting reagent (LR) = 0.09285
mols H2O produced IF we used NaOH as the LR = 0.2740
In LR problem we always produce the LEAST amount possible and have an excess of the other reagent; i.e., we will have an excess of NaOH.
To determine the number of moles of water produced in this neutralization reaction, we first need to identify the balanced chemical equation for the reaction between hydrochloric acid (HCl) and sodium hydroxide (NaOH).
The balanced chemical equation for the neutralization reaction between HCl and NaOH is:
HCl + NaOH → NaCl + H2O
From the balanced equation, we can see that 1 mole of HCl reacts with 1 mole of NaOH to produce 1 mole of water.
Now, let's calculate the number of moles of HCl and NaOH in the given solutions:
The volume (in liters) of the HCl solution is 100.0 mL, which is equal to 0.1000 L.
To calculate the number of moles of HCl, we will use the formula:
moles = concentration (M) x volume (L)
The concentration of HCl is 0.9285 M, and the volume is 0.1000 L.
Therefore, moles of HCl = 0.9285 M x 0.1000 L = 0.09285 mol
Similarly, for the NaOH solution:
The volume (in liters) of the NaOH solution is 150.0 mL, which is equal to 0.1500 L.
To calculate the number of moles of NaOH, we will use the formula:
moles = concentration (M) x volume (L)
The concentration of NaOH is 1.8269 M, and the volume is 0.1500 L.
Therefore, moles of NaOH = 1.8269 M x 0.1500 L = 0.2739 mol
Since the balanced equation states that 1 mole of HCl reacts with 1 mole of NaOH to produce 1 mole of water, the number of moles of water produced in this reaction is equal to the smaller of the two amounts of reactants.
In this case, the number of moles of water produced is 0.09285 mol.
Therefore, 0.09285 moles of water are produced by this neutralization.