Question

20. Identify the neutralization reaction. Classify all of the reactions shown below.
a) CH4 (g) + 2 O2 (g) → CO2 (g) + 2 H2O (l)
b) MgBr2 (aq) + Cl2 (g) → MgCl2 (aq) + Br2 (l)
c) HCl (aq) + NaOH (aq) → NaCl (aq) + H2O (l)
d) BaCl2 (aq) + 2 AgNO3 (aq) → 2 AgCl (s) + Ba(NO3)2 (aq)

23) Calculate the mass and the number of the hydroxide ions in Ca(OH)2 required to react with the acetic acid, CH3CO2H, in 25.0 mL of a solution having a density of 1.065 g/mL and containing 58.0% acetic acid by mass. Be sure to include a balanced chemical equation.

24) A 0.652-gram sample of pure strontium halide reacts with excess H2SO4, and the solid strontium sulfate formed is separated, dried and found to weigh 0.755 grams. What is the formula of the original halide?

Answer 1

20)
a) CH4 (g) + 2 O2 (g) → CO2 (g) + 2 H2O (l)...................> Combustion reaction.
b) MgBr2 (aq) + Cl2 (g) → MgCl2 (aq) + Br2 (l)................> single replacement reaction.
c) HCl (aq) + NaOH (aq) → NaCl (aq) + H2O (l).....................> Neutralization reaction as NaOH is base and HCl is acid.
d) BaCl2 (aq) + 2 AgNO3 (aq) → 2 AgCl (s) + Ba(NO3)2 (aq)................> Double replacement reaction.

23) Calculate the mass and the number of the hydroxide ions in Ca(OH)2 required to react with the acetic acid, CH3CO2H, in 25.0 mL of a solution having a density of 1.065 g/mL and containing 58.0% acetic acid by mass. Be sure to include a balanced chemical equation.

balanced reaction is

Ba(OH)2 + 2 CH3COOH ......................> Ba(CH3COO)2 + 2 H2O

mass of solution = 25.0 ml * 1.065 g/mL = 26.625 g

mass of acetic acid = 26.625 * 58.0 / 100 = 15.44 g

mole of acetic acid = 15.44 g / 60 g / mole = 0.257 mole.

number of hydroxide ion = 0.257 * 6.023 * 10^23 = 1.55 * 10^23

mass of hydroxide ion = mole * molar mass = 0.257 mole * 17 g / mole = 4.37 g

24) A 0.652-gram sample of pure strontium halide reacts with excess H2SO4, and the solid strontium sulfate formed is separated, dried and found to weigh 0.755 grams. What is the formula of the original halide?

0.755 g SrSO4 = mass / molar mass = 0.755 g / 183.68 g/mole = 4.11 * 10^-3 mole.

thus

mole of strontium halide = 4.11 * 10^-3 mole.

molar mass of strontium halide = 0.652 g / 4.11 * 10^-3 mole = 158.62 g / mole.

mass of halide = 158.62 - 87.62 = 71.0

thus

formula of the original halide is SrCl₂.