A 70% solution of HNO3 with a volume of 202.7 (density 1.421 g / cm3) was used for a 200 g sample of technical

| February 9, 2021 | education

A 70% solution of HNO3 with a volume of 202.7 (density 1.421 g / cm3) was used for a 200 g sample of technical calcium carbonate. Find the percentage of impurities in the sample.

ω (%) = m1 / m * 100%
In our example, the volume and density are given, which means we can calculate the mass of the solution
Density of solution = mass of solution / volume of solution
Ρ = m / v Hence m = Ρ * v
m solution = 1.421 * 202.7 = 288.0367 grams
m1 = ω (%) * m = 70 * 288.0367 / 100% = 201.62569
CaCO3 + 2HNO3 = Ca (NO3) 2 + CO2 + H2O
Determine the mole of HNO3 (n)
n = m1 / M where M is the molecular weight of HNO3
M = 1 + 14+ 3 * 16 = 63
n = 201.62569 / 63 = 3.2
1 mol of CaCO3 reacts with 2 mol of HNO3
so 3.2 will be 3.2 * 1/2 = 1.6
M (CaCO3) = M (Ca) + M (C) + 3M (O) = 40 + 12 + 3 * 16 = 100
m CaCO3 = 1.6 * M CaCO3 = 1.6 * 100 = 160 Grams
m impurities = 200-160 = 40 grams
percentage of impurities in the sample = 40/200 * 100% = 20%



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.