If I combined 0.202 mol of calcium hydroxide with 75.0 mL of 0.500 M HCl: Ca(OH)₂ + 2 HCl → CaCl₂ + 2 H₂OHow many moles of calcium chloride would be formed?

To solve this problem, we need to use stoichiometry, which is a method in chemistry for calculating amounts in reactions.

Step 1: Identify the balanced chemical equation The balanced chemical equation is given as: Ca(OH)₂ + 2 HCl → CaCl₂ + 2 H₂O

Step 2: Convert the volume of HCl to moles We know that Molarity (M) = moles/volume(L). So, we can find the moles of HCl by multiplying the molarity by the volume (in liters). 75.0 mL = 0.075 L So, moles of HCl = Molarity * Volume = 0.500 M * 0.075 L = 0.0375 mol

Step 3: Use the stoichiometry of the reaction to find the moles of CaCl₂ From the balanced chemical equation, we can see that 2 moles of HCl react with 1 mole of Ca(OH)₂ to produce 1 mole of CaCl₂. So, the moles of CaCl₂ produced will be the same as the moles of Ca(OH)₂ reacted.

Step 4: Identify the limiting reactant We have 0.202 mol of Ca(OH)₂ and 0.0375 mol of HCl. According to the stoichiometry of the reaction, 1 mole of Ca(OH)₂ can react with 2 moles of HCl. But we have less than 0.202*2 = 0.404 moles of HCl. So, HCl is the limiting reactant.

Step 5: Calculate the moles of CaCl₂ formed Since HCl is the limiting reactant, the moles of CaCl₂ formed will be half the moles of HCl (because 2 moles of HCl produce 1 mole of CaCl₂). So, moles of CaCl₂ = 0.0375 mol / 2 = 0.01875 mol

So, 0.01875 moles of calcium chloride would be formed.