Question

Визначити масу осаду, що утворится при взаємодії 148г 10% Сa(OH)2 з CO2

Answer 1

Дано:

m(Ca(OH)2) = 148 г

w(Ca(OH)2) = 10%=0,1

Найти:

m(CaCO3) - ?

Решение:

Ca(OH)2 + CO2 => CaCO3 + H2O

m(в-ва)= w×m(р-ра)

m(Ca(OH)2) = 0,1 × 148 = 14,8 г

n=m/M

M(Ca(OH)2) = 1 × 40 + 2 × 16 + 2 × 1 = 74 г/моль

n(Ca(OH)2) = 14,8/74 = 0,2 моль

n(CaCO3) = n(Ca(OH)2) = 0,2 моль

m=n×M

M(CaCO3) = 1 × 40 + 1 × 12 + 3 × 16 = 100 г/моль

m(CaCO3) = 0,2 × 100 = 20 г

Ответ: 20 г

Answer 2

Calculation of the Mass of Precipitate Formed

To determine the mass of the precipitate formed when 148g of 10% Ca(OH)2 reacts with CO2, we need to consider the balanced chemical equation for the reaction between Ca(OH)2 and CO2.

The balanced chemical equation for the reaction is as follows: Ca(OH)2 + CO2 -> CaCO3 + H2O

From the equation, we can see that 1 mole of Ca(OH)2 reacts with 1 mole of CO2 to produce 1 mole of CaCO3 and 1 mole of H2O.

To calculate the mass of the precipitate formed, we need to determine the number of moles of Ca(OH)2 and CO2 involved in the reaction.

1. Calculate the number of moles of Ca(OH)2: - The molar mass of Ca(OH)2 is 74.09 g/mol (40.08 g/mol for Ca + 2 * 1.01 g/mol for H + 16.00 g/mol for O). - We have 148g of a 10% solution of Ca(OH)2, which means there are 10g of Ca(OH)2 in 100g of the solution. - Therefore, the mass of Ca(OH)2 in the solution is (10/100) * 148g = 14.8g. - The number of moles of Ca(OH)2 can be calculated using the formula: moles = mass / molar mass. - So, the number of moles of Ca(OH)2 is 14.8g / 74.09 g/mol = 0.1996 mol.

2. Calculate the number of moles of CO2: - The molar mass of CO2 is 44.01 g/mol (12.01 g/mol for C + 2 * 16.00 g/mol for O). - The balanced chemical equation tells us that 1 mole of Ca(OH)2 reacts with 1 mole of CO2. - Therefore, the number of moles of CO2 is also 0.1996 mol.

Since the stoichiometry of the reaction is 1:1 for Ca(OH)2 and CO2, we can conclude that 0.1996 mol of CaCO3 will be formed.

To calculate the mass of the precipitate (CaCO3), we can use the formula: mass = moles * molar mass.

The molar mass of CaCO3 is 100.09 g/mol (40.08 g/mol for Ca + 12.01 g/mol for C + 3 * 16.00 g/mol for O).

Therefore, the mass of the precipitate formed is 0.1996 mol * 100.09 g/mol = 19.96 g.

The mass of the precipitate formed when 148g of 10% Ca(OH)2 reacts with CO2 is approximately 19.96 grams.