How much limestone with a calcium carbonate content of 70% is required to completely neutralize
| March 18, 2021 | education
How much limestone with a calcium carbonate content of 70% is required to completely neutralize 10 liters of sulfuric acid with a concentration of 5 g / l.
M (H2SO4) = 5 g. * 10 l. = 50 g.
СaCO3 + H2SO4 = CaSO4 + CO2 + H2O
n (CaCO3) = n (H2SO4)
n (H2SO4) = 50 g / 98 g / mol = 0.51 mol
m₁ (CaCO3) = n * M = 0.51 * 100 g / mol = 51 g.
m₂ (CaCO3) = 51g. * 100%: 70% = 72.86 g is the mass of limestone.
One of the components of a person's success in our time is receiving modern high-quality education, mastering the knowledge, skills and abilities necessary for life in society. A person today needs to study almost all his life, mastering everything new and new, acquiring the necessary professional qualities.
