In the reaction Ca(OH)2 + 2HCL -> CaCl2 + 2H2O: How many kilograms of H2O will be produced when 0.0948 Kg of Ca(OH)2 react with excess hydrochloric acid?
Isn't this a regular stoichiometry problem?
yes it is but i need help in how to solve it.
Here is how all of them are worked.
1. Write and balance the equation.
Ca(OH)2 + 2HCl ==> CaCl2 + 2H2O
2. Convert what you have to mols.
mols Ca(OH)2 = grams/molar mass = ? You will need to convert kg to g first.
3. Using the coefficients in the balanced equation, convert mols Ca(OH)2 to mols H2O produced. In this case that is ?mols Ca(OH)2 x (2 mols H2O/1 mol Ca(OH)2) = ?mols Ca(OH)2 x 2/1 = ?
4. Then convert mols H2O to grams. g = mols x molar mass = ? and convert to kg.
Post your work if you get stuck or explain in detail what you don't understand.
To determine the amount of H2O produced in the reaction, we need to use stoichiometry, which is the relationship between the amounts of reactants and products in a chemical reaction. In this case, we can use the balanced chemical equation to determine the ratio of Ca(OH)2 to H2O produced.
The balanced chemical equation is:
Ca(OH)2 + 2HCl -> CaCl2 + 2H2O
From the equation, we can see that 1 mole of Ca(OH)2 produces 2 moles of H2O. To convert from moles to kilograms, we need to know the molar mass of Ca(OH)2 and H2O.
The molar mass of Ca(OH)2:
Ca: 40.08 g/mol
O: 16.00 g/mol
H: 1.01 g/mol
So, the molar mass of Ca(OH)2 is:
40.08 + 2*(16.00 + 1.01) = 74.14 g/mol
Now, we can calculate the number of moles of Ca(OH)2:
moles of Ca(OH)2 = mass of Ca(OH)2 / molar mass of Ca(OH)2
moles of Ca(OH)2 = 0.0948 kg / 74.14 g/mol
Next, we use the stoichiometry from the balanced equation to calculate the number of moles of H2O produced:
moles of H2O = moles of Ca(OH)2 * (2 moles of H2O / 1 mole of Ca(OH)2)
Finally, we convert the moles of H2O to kilograms:
mass of H2O = moles of H2O * molar mass of H2O
So, by following these steps, we can determine the mass of H2O produced when 0.0948 kg of Ca(OH)2 reacts with excess hydrochloric acid.