If 45.0 mL of 0.25 M HCl is required to completely neutralize 25.0 mL of NH3, what is the concentration of the NH3 solution? Show all of the work needed to solve this problem.

HCl + NH3 yields NH4Cl

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cristisp93 cristisp93 · Answer 1 · 2019-07-24T09:28:04+02:00

molar concentration = number of moles / volume

number of moles = molar concentration × volume

Number of moles of HCl = 0.25 × 45 = 11.25 mmoles

From the reaction:

11.25 mmoles of HCl will neutralize 11.25 mmoles of NH₃

Molar concentration of NH₃ solution = 11.25 / 25 = 0.45 M

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donotbeidle donotbeidle · Answer 2 · 2019-10-03T18:42:03+02:00

Answer:

The balanced chemical equation is

[tex]1HCl+1NH_3>1NH_4 Cl[/tex]

The conversions are

Molarity of HCl and volume gives us moles HCl

Moles HCl to moles [tex]NH_3[/tex] (using mole ratio 1 : 1)

Moles [tex]NH_3[/tex] to Molarity [tex]NH_3[/tex]

Molarity HCl = (moles solute HCl) / (volume of solution in L)

Rearranging the formula

We get moles HCl = Molarity × volume

[tex]=0.25M \times 45.0mL = 0.25 mol/L \times 0.045L = 0.01125 mol HCl[/tex]

moles HCl = moles [tex]NH_3[/tex] = 0.01125 mol [tex]NH_3[/tex]

Molarity [tex]NH_3[/tex] =(moles solute [tex]NH_3[/tex]) / (volume of solution in L)

[tex]=\frac {(0.01125 mol )}{25.0mL} = \frac {(0.01125 mol )}{0.025L}[/tex]

=0.450 mol/L or M (Answer)