How many grams of CaO (56.08) are required for complete reaction with HCl (36.46) in 275 mL of a 5.90 M HCl solution? CaO + 2HCl --> CaCl2 + H2O.
mols HCl = M x L = 5.90M x 0.275L = ?
Look at the coefficients in the balanced equation to find mols CaO = mols HCl x 1/2.
Then g CaO = mols CaO x molar mass CaO
12.2
To find the number of grams of CaO required for complete reaction with HCl, we'll need to follow these steps:
1. Calculate the moles of HCl in the solution:
Moles of HCl = Molarity of HCl × Volume of HCl solution in liters
= 5.90 M × 0.275 L
= 1.6175 moles
2. Determine the stoichiometric ratio between HCl and CaO from the balanced chemical equation:
It tells us that 1 mole of CaO reacts with 2 moles of HCl.
3. Use the stoichiometry to find the moles of CaO required:
Moles of CaO = (Moles of HCl ÷ 2)
= (1.6175 moles ÷ 2)
= 0.80875 moles
4. Calculate the mass of CaO using its molar mass (56.08 g/mol):
Mass of CaO = Moles of CaO × Molar mass of CaO
= 0.80875 moles × 56.08 g/mol
= 45.4072 grams
Therefore, approximately 45.41 grams of CaO are required for the complete reaction with HCl.