What mass of CaCo3 is required to obtain Co2 volume = 11.2 liters?
August 30, 2021
Let’s write down the solution:
1. According to the condition of the problem, we compose the equation:
X g -? V = 11.2 liters.
CaCO3 = CaO + CO2 – decomposition, carbon monoxide obtained (4);
2. Let’s make the calculations:
M (CaCO3) = 70 g / mol.
M (CO2) = 44 g / mol.
1 mol of gas at normal level – 22.4 liters.
X mol (CO2) -11.2 l. hence, X mol (CO2) = 1 * 11.2 / 22.4 = 0.5 mol.
Y (CaCO3) = 0.5 mol since the amount of substances is 1 mol.
4. Find the mass of the original substance:
m (CaCO3) = Y * M = 0.5 * 70 = 35 g.
Answer: The mass of calcium carbonate is 35 g.
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.