CaO+2HCl=CaCl₂+H₂O
n(CaO)=0.2 Моль
m(CaCl₂) - ?

Предыдущий
вопрос Следующий
вопрос

To find the mass of CaCl₂ produced when 0.2 moles of CaO react with excess HCl, we first need to determine the stoichiometry of the reaction.

From the balanced chemical equation:
1 mole of CaO reacts with 2 moles of HCl to produce 1 mole of CaCl₂ and 1 mole of H₂O.

Given that the amount of CaO is 0.2 moles, this means that it will react with 0.2 * 2 = 0.4 moles of HCl to produce 0.2 moles of CaCl₂.

Now, let's calculate the molar mass of CaCl₂:
Ca = 40.08 g/mol
Cl = 35.45 g/mol (x2 because there are 2 atoms of Cl)
Molar mass of CaCl₂ = 40.08 + 2*35.45 = 110.98 g/mol

Now, we can calculate the mass of 0.2 moles of CaCl₂:
Mass = moles molar mass
Mass = 0.2 110.98 = 22.196 g

Therefore, when 0.2 moles of CaO react with excess HCl, 22.196 grams of CaCl₂ will be produced.