how many grams of CaO are required for complete reaction with the HCl in 275mL of a 0.523 M HCl?
write the balanced equation:
CaO + 2HCl >> CaCl2 + H2O
Note then that it takes half as many moles of CaO as it does HCl.
Moles HCl: .275*.523
moles CaO: half that.
grams CaO: molesCaO*molmassCaO
Assistance needed.
Please type your subject in the School Subject box. Any other words, including obscure abbreviations, are likely to delay responses from a teacher who knows that subject well.
To calculate the number of grams of CaO required for a complete reaction with HCl, we need to use stoichiometry and the balanced chemical equation.
The balanced chemical equation for the reaction between CaO and HCl is:
CaO + 2HCl -> CaCl2 + H2O
From the equation, we can see that 1 mole of CaO reacts with 2 moles of HCl.
First, let's calculate the number of moles of HCl present in the solution:
Molarity (M) = moles of solute / volume of solution (in liters)
Given:
Molarity of HCl (M) = 0.523 M
Volume of HCl solution = 275 mL = 0.275 L
moles of HCl = Molarity × volume
moles of HCl = 0.523 M × 0.275 L = 0.143275 mol
Since the stoichiometric ratio between CaO and HCl is 1:2, we have:
moles of CaO = moles of HCl / 2
moles of CaO = 0.143275 mol / 2 = 0.0716375 mol
Finally, to calculate the mass of CaO required, we need to use its molar mass.
The molar mass of CaO is:
Ca: 40.08 g/mol
O: 16.00 g/mol
Molar mass of CaO = 40.08 g/mol + 16.00 g/mol = 56.08 g/mol
Mass of CaO = moles of CaO × molar mass of CaO
Mass of CaO = 0.0716375 mol × 56.08 g/mol = 4.01929 g
Therefore, approximately 4.02 grams of CaO are required for a complete reaction with the HCl solution.
To determine the amount of CaO required for the reaction, you can use the stoichiometry of the balanced chemical equation between CaO and HCl, as well as the given concentration and volume of HCl.
The balanced chemical equation between CaO and HCl is:
CaO + 2HCl → CaCl2 + H2O
From the equation, we can see that 1 mole of CaO reacts with 2 moles of HCl.
First, we need to calculate the number of moles of HCl present in the given volume of the solution. We can use the equation:
moles = concentration (M) × volume (L)
Given:
Concentration of HCl (C) = 0.523 M
Volume of HCl (V) = 275 mL = 0.275 L
Now, we can calculate the moles of HCl:
moles of HCl = 0.523 M × 0.275 L = 0.143275 moles
Since the stoichiometric ratio between CaO and HCl is 1:2, the number of moles of CaO required will be half of the moles of HCl:
moles of CaO = 0.143275 moles ÷ 2 = 0.0716375 moles
Finally, we can calculate the mass of CaO using its molar mass:
Molar mass of CaO = 40.08 g/mol (calcium: 40.08 g/mol + oxygen: 16.00 g/mol)
Mass of CaO = moles of CaO × molar mass of CaO
Mass of CaO = 0.0716375 moles × 56.08 g/mol ≈ 4.02 grams
Therefore, to completely react with the HCl in 275 mL of a 0.523 M HCl solution, approximately 4.02 grams of CaO is required.