This is an acid base reaction and the chemical equation for the above reaction is as follows;
KOH + HClO₄ ---> KClO₄ + H₂O
the stoichiometry of acid to base is 1:1
KOH is a strong base and HClO₄ is a strong acid therefore they both ionize completely into their respective ions
Number of KOH moles - 0.723 M/1000 mL/L x 25.0 mL = 0.018 mol
Number of HClO₄ moles - 0.273 M/1000 mL/L x 50 mL = 0.013 mol
since acid and base react completely, 0.013 mol of acid reacts with 0.013 mol of base.
The excess base remaining is - 0.018 - 0.013 = 0.005 mol
total volume of solution = 25.0 mL + 50.0 mL = 75.0 mL
[OH⁻] = 0.005 mol/0.075 L = 0.067 M
pOH = -log[OH⁻]
pOH = -log(0.067 M)
pOH = 1.17
pOH + pH = 14
Therefore pH = 14 - 1.17 = 12.83
by knowing pH we can calculate the [H₃O⁺]
pH = -log [H₃O⁺]
[H₃O⁺] = antilog[-12.83]
[H₃O⁺]= 1.47 x 10⁻¹³ M
Answer:
2.0 moles
Explanation:
I hope this helps you a little bit at least the answer is 2.0 but if you want to review more stuff check the photos
Volume of osmium = 1.01(0.223)(0.648) = 0.14595 cm3
Density = mass / volume
So density x volume = mass of osmium
22.6 x 0.14595 = 3.29845 g
Answer:
5 mol.
Explanation:
Equation of the reaction
2SO2 + 2H2O + O2 --> 2H2SO4
By stoichiometry, 2 moles of SO2 reacted with 2 moles of water and 1 mole of O2 to give 2 mole of sulphuric acid.
Number of moles:
5.0 mol SO2
4.0 mol O2
20.0 mol H2O
Calculating the limiting reagent,
5 mol of SO2 * 1 mol of O2/2 mol of SO2
= 2.5 mol of O2(4 mol of O2 is present)
5 mol of SO2 * 2 mol of H2O/2 mol of SO2
= 5 mol of H2O(20 mol of H2O)
SO2 is the limiting reagent.
Therefore, number of moles of H2SO4 = 5 mol of SO2 * 2 mol of H2SO4/2 mol of SO2
= 5 mol of H2SO4.
