Question

Дано m CaO-28г и m HCl-73г найти m(соли)

Answer 1

Вот, держи решение. Надеюсь, помогла :)

Answer 2

To find the mass of the salt (CaCl2) formed when 28g of CaO reacts with 73g of HCl, we first need to write and balance the chemical equation for the reaction between calcium oxide (CaO) and hydrochloric acid (HCl):

CaO + 2 HCl -> CaCl2 + H2O

The balanced equation shows that 1 mole of CaO reacts with 2 moles of HCl to produce 1 mole of CaCl2 and 1 mole of water (H2O).

Next, we need to calculate the number of moles of each reactant:

1. Moles of CaO: Molar mass of CaO (Ca: 40.08 g/mol, O: 16.00 g/mol) = 40.08 + 16.00 = 56.08 g/mol Moles of CaO = Mass of CaO / Molar mass of CaO = 28 g / 56.08 g/mol ≈ 0.499 moles

2. Moles of HCl: Molar mass of HCl (H: 1.01 g/mol, Cl: 35.45 g/mol) = 1.01 + 35.45 = 36.46 g/mol Moles of HCl = Mass of HCl / Molar mass of HCl = 73 g / 36.46 g/mol ≈ 2.002 moles

Since 1 mole of CaO reacts with 2 moles of HCl, we have an excess of CaO. The limiting reagent is HCl, and it will be completely consumed in the reaction.

Now, we can determine the moles of CaCl2 formed based on the balanced equation:

Moles of CaCl2 = Moles of HCl (since 1 mole of CaO produces 1 mole of CaCl2)

Moles of CaCl2 = 2.002 moles

Finally, to find the mass of CaCl2 (m(salt)), we use its molar mass:

Molar mass of CaCl2 (Ca: 40.08 g/mol, Cl: 35.45 g/mol x 2) = 40.08 + 70.90 = 110.98 g/mol

Mass of CaCl2 = Moles of CaCl2 x Molar mass of CaCl2 Mass of CaCl2 = 2.002 moles x 110.98 g/mol ≈ 222.18 g

Therefore, the mass of the salt (CaCl2) formed in the reaction is approximately 222.18 grams.