Question

Вычислите массу осадка который образуется при сливании 3 г NaOH и 4 г AlCl3

Answer 1

AlCl3 + 3NaOH (недостаток)=Al(OH)3 + 3NaCl
n=0.03  n=0.025
n(NaOH)=3/40=0.0075 моль
n(Al(OH)3)=0.075*1/3=0.025 моль
m(Al(OH)3)=0.025*78=1,95 г

Answer 2

Calculating the Mass of Precipitate Formed

To calculate the mass of the precipitate formed when 3g of NaOH and 4g of AlCl3 are mixed, we can use the concept of a chemical reaction between the two compounds. The balanced chemical equation for the reaction between NaOH and AlCl3 is:

2NaOH + AlCl3 -> 2NaCl + Al(OH)3

From the balanced chemical equation, we can see that 2 moles of NaOH react with 1 mole of AlCl3 to produce 2 moles of NaCl and 1 mole of Al(OH)3.

Using the molar masses of the compounds: - Molar mass of NaOH = 40 g/mol - Molar mass of AlCl3 = 133.34 g/mol

We can calculate the moles of NaOH and AlCl3 and determine the limiting reactant to find the mass of the precipitate formed.

First, let's calculate the moles of NaOH and AlCl3: - Moles of NaOH = 3g / 40 g/mol = 0.075 moles - Moles of AlCl3 = 4g / 133.34 g/mol = 0.03 moles

Since the stoichiometry of the reaction is 2:1 for NaOH and AlCl3, AlCl3 is the limiting reactant. Therefore, all 0.03 moles of AlCl3 will react with 0.06 moles of NaOH to form the precipitate.

The molar mass of Al(OH)3 is 78 g/mol. Therefore, the mass of the precipitate formed can be calculated as: - Mass of Al(OH)3 = 0.03 moles * 78 g/mol = 2.34g

The mass of the precipitate formed when 3g of NaOH and 4g of AlCl3 are mixed is 2.34g.