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Answer: D. 19.9 g hydrogen remains.
Explanation:
To calculate the moles, we use the equation:
a) moles of
b) moles of
According to stoichiometry :
1 mole of require 1 mole of
Thus 0.0787 moles of require=
of
Thus is the limiting reagent as it limits the formation of product and
acts as the excess reagent. (10.0-0.0787)= 9.92 moles of
are left unreacted.
Mass of
Thus 19.9 g of remains unreacted.
Answer:
pH ≅ 4.80
Explanation:
Given that:
the volume of HN₃ = 25 mL = 0.025 L
Molarity of HN₃ = 0.150 M
number of moles of HN₃ = 0.025 × 0.150
number of moles of HN₃ = 0.00375 mol
Molarity of NaOH = 0.150 M
the volume of NaOH = 13.3 mL = 0.0133
number of moles of NaOH = 0.0133× 0.150
number of moles of NaOH = 0.001995 mol
The chemical equation for the reaction of this process can be written as:
1 mole of hydrazoic acid react with 1 mole of hydroxide to give nitride ion and water
thus the new number of moles of HN₃ = 0.00375 - 0.001995 = 0.001755 mol
Total volume used in the reaction = 0.025 + 0.0133 = 0.0383 L
Concentration of =
= 0.0458 M
Concentration of =
= 0.0521 M
GIven that :
Ka =
Thus; it's pKa = 4.72
pH ≅ 4.80