Question

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

Answer 1

M(CaCl2) = n * M= 0,2 моль * 111 = 22.2 грамм

Answer 2

To find the mass of calcium chloride (CaCl₂) formed when 0.2 moles of calcium oxide (CaO) reacts with excess hydrochloric acid (HCl), we need to use the stoichiometry of the balanced chemical equation.

The balanced chemical equation is:

CaO + 2HCl → CaCl₂ + H₂O

From the equation, we can see that 1 mole of CaO reacts with 2 moles of HCl to produce 1 mole of CaCl₂.

Given that n(CaO) = 0.2 moles, we can use this information to calculate the moles of CaCl₂ formed:

Moles of CaCl₂ = n(CaO) = 0.2 moles

Now, to find the molar mass of CaCl₂, we sum the atomic masses of its constituent elements:

CaCl₂ molar mass = (40.08 g/mol for Ca) + (2 * 35.45 g/mol for Cl) ≈ 110.98 g/mol

Finally, to find the mass (m) of CaCl₂ formed, we can use the formula:

m = moles of CaCl₂ * molar mass of CaCl₂

m = 0.2 moles * 110.98 g/mol ≈ 22.196 g

So, approximately 22.196 grams of calcium chloride (CaCl₂) will be formed when 0.2 moles of calcium oxide (CaO) reacts with excess hydrochloric acid (HCl).