Question

Consider the following reaction:

2HCl + CaCO₃→ CaCl₂ + H₂O + CO₂
How many mols of calcium chloride can be produced if you begin with 8.4 mL of 0.62 M HCl and have an excess amount of calcium carbonate?

Answer 1

Approximately 0.289 grams of calcium chloride can be produced.

Step-by-step explanation:

In the given chemical equation, the stoichiometric coefficient in front of CaCl₂ is 1. This means that for every 2 moles of HCl, 1 mole of CaCl₂ is produced.

First, we need to calculate the number moles of HCl from the given volume and molarity:

moles of HCl = volume (in L) × molarity

moles of HCl = 8.4 mL × (1 L / 1000 mL) × 0.62 M

moles of HCl = 0.005208 moles

According to the stoichiometry of the reaction, the number moles of CaCl₂ produced will be half of the number of moles of HCl:

moles of CaCl₂ = 0.005208 moles / 2 = 0.002604 moles

To convert the number of moles of CaCl₂ to grams, we need to use the molar mass of CaCl₂:

molar mass of CaCl₂ = 40.08 g/mol + 35.45 g/mol × 2 = 110.98 g/mol

mass of CaCl₂ = moles of CaCl₂ × molar mass

mass of CaCl₂ = 0.002604 moles × 110.98 g/mol = 0.289 g

Therefore, approximately 0.289 grams of calcium chloride can be produced.