The volume of H2S needed to neutralize 45mL of NH2- is 0.053 L
What do you mean by neutralization reaction?A neutralization reaction, in which an acid and a base react to form water and salt.
The reaction between H2S and NH2- is a neutralization reaction. The balanced equation for the reaction is:
H2S(aq) + NH2-(aq) → NH4HS(aq)
We know that the concentration of H2S is 1.2 M and the concentration of NH2- is 0.80 M and we are trying to find the volume of H2S needed to neutralize 45mL of NH2-.
The number of moles of H2S can be calculated by using the formula:
moles = molarity x volume
The volume of H2S needed to neutralize 45mL of NH2- is equal to the number of moles of NH2- in 45mL, divided by the concentration of H2S.
moles of NH2- = 0.080 L * 0.80 mol/L = 0.064 mol
volume = moles of NH2- / molarity of H2S
volume = 0.064 mol / 1.2 mol/L = 0.053 L
Therefore, the volume of H2S needed to neutralize 45mL of NH2- is 0.053 L.
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